Quadratic Equations

JEE Main Mathematics notes. RelaxPrep is not affiliated with the NTA or the IITs.

Revision sheet

A quadratic $ax^2+bx+c=0$ has $a\neq0$, roots $\frac{-b\pm\sqrt{D}}{2a}$ and discriminant $D=b^2-4ac$.
$D>0$: two distinct real roots; $D=0$: equal real roots; $D<0$: non-real conjugate roots (real coefficients).
Sum of roots $=-\frac ba$, product $=\frac ca$; the equation with roots $\alpha,\beta$ is $x^2-(\alpha+\beta)x+\alpha\beta=0$.
Symmetric expressions are found from $S=\alpha+\beta$ and $P=\alpha\beta$: $\alpha^2+\beta^2=S^2-2P$, $\alpha^3+\beta^3=S^3-3SP$.
If $a+b+c=0$ the roots are $1$ and $\frac ca$; if $a-b+c=0$ they are $-1$ and $-\frac ca$.
Vertex $\left(-\frac b{2a},-\frac D{4a}\right)$; outside the roots $f$ has the sign of $a$, between them the opposite sign.
Always positive: $a>0$ and $D<0$. $k$ between the roots: $a\,f(k)<0$.
Both roots greater than $k$: $D\ge0$, $-\frac b{2a}>k$, $a\,f(k)>0$.
Wavy curve: sign flips at odd-power roots only; denominator roots are excluded.
Cubic roots: $\sigma_1=-\frac ba$, $\sigma_2=\frac ca$, $\sigma_3=-\frac da$ (alternating signs).

Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Vieta: $\alpha+\beta=-\dfrac ba,\qquad \alpha\beta=\dfrac ca$
Common root condition: $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$
Extreme value: $-\dfrac{D}{4a}\ \text{at}\ x=-\dfrac{b}{2a}$
Real quadratic with root $p+iq$: $x^2-2px+\left(p^2+q^2\right)=0$
Cubic from roots: $x^3-\sigma_1x^2+\sigma_2x-\sigma_3=0$

Topics

Quadratic equation, roots and the quadratic formula

A quadratic equation in $x$ is one that can be written as $ax^2+bx+c=0$ with $a\neq 0$. A number $r$ is a root (or solution) when substituting $x=r$ makes the left side zero. Over the complex numbers every quadratic has exactly two roots, counted with multiplicity; over the reals it can have two, one repeated, or none.

The standard route to the roots is completing the square. Divide by $a$, add and subtract $\left(\frac{b}{2a}\right)^2$, and the equation becomes $\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$. Taking square roots gives the quadratic formula below. Factorising is the faster route when the roots are rational and small, for example $2x^2-7x+3=(2x-1)(x-3)$ with roots $\frac12$ and $3$.

A polynomial of degree two cannot have three distinct roots unless all its coefficients are zero. So if $ax^2+bx+c=0$ holds for three different values of $x$, then $a=b=c=0$. This identity-versus-equation distinction is used to compare coefficients when two quadratic expressions are claimed equal for all $x$.

Key points

The equation $ax^2+bx+c=0$ needs $a\neq 0$; if $a=0$ it is linear (or degenerate).
The two roots are $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
Completing the square rewrites $ax^2+bx+c$ as $a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}$.
If a quadratic expression vanishes for three distinct values of $x$, all its coefficients are zero.

Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Completed square: $ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a}$

Common traps

Dividing both sides by an expression containing $x$ (for example cancelling a common factor $x-1$) can lose a root; move everything to one side and factorise instead.
Squaring both sides, or multiplying by an expression in $x$, can add roots that do not satisfy the original equation. Always substitute back when the original equation had radicals, denominators or logarithms.
If the leading coefficient contains a parameter, such as $(k-2)x^2+8x+(k+4)=0$, the case where it vanishes must be treated separately: the equation becomes linear there.

Shortcuts

Roots at x = 1 or x = -1 by coefficient sum. If $a+b+c=0$, then $x=1$ is a root and the other root is $\frac{c}{a}$. If $a-b+c=0$, then $x=-1$ is a root and the other is $-\frac{c}{a}$. Example: for $5x^2-8x+3$ we have $5-8+3=0$, so the roots are $1$ and $\frac35$.

Discriminant and nature of roots

The quantity $D=b^2-4ac$ sits under the square root in the quadratic formula, so its sign decides what kind of roots appear. For real coefficients: $D>0$ gives two distinct real roots, $D=0$ gives one repeated real root $x=-\frac{b}{2a}$, and $D<0$ gives two non-real roots that are complex conjugates of each other.

If the coefficients are rational, the square root of $D$ decides rationality. When $D$ is a perfect square of a rational number the roots are rational. When $D>0$ but not a perfect square the roots are irrational, and they come as a conjugate pair $p\pm\sqrt{q}$ with $p,q$ rational. For integer coefficients the same statement holds with $D$ a perfect square integer.

A useful sign fact: if $a$ and $c$ have opposite signs then $ac<0$, so $D=b^2-4ac>0$ whatever $b$ is, and the roots are real and distinct. Another standard identity: for the equation $(x-A)(x-B)+(x-B)(x-C)+(x-C)(x-A)=0$, the discriminant equals $2\left[(A-B)^2+(B-C)^2+(C-A)^2\right]$, which is non-negative, so its roots are always real, and distinct unless $A=B=C$.

Parameter problems usually reduce to an inequality in the parameter. For $(k-2)x^2+8x+(k+4)=0$ to have equal roots we need $64-4(k-2)(k+4)=0$, that is $k^2+2k-24=0$, so $k=4$ or $k=-6$ (both keep $k\neq 2$).

Key points

$D>0$: two distinct real roots. $D=0$: two equal real roots. $D<0$: two conjugate non-real roots (real coefficients).
Rational coefficients and $D$ a perfect square: both roots rational.
Rational coefficients, $D>0$ and not a perfect square: roots $p\pm\sqrt q$, both irrational.
$ac<0$ forces $D>0$.
The conjugate-pair statements need real (or rational) coefficients; they fail when the coefficients are irrational or complex.

Discriminant: $D=b^2-4ac$
Repeated root when $D=0$: $x=-\dfrac{b}{2a}$

Common traps

$D=0$ means a single repeated root (counted twice), not that the equation has no root.
Irrational roots occur in conjugate pairs only for rational coefficients. The equation $x^2-(1+\sqrt2)x+\sqrt2=0$ has roots $1$ and $\sqrt2$: one rational and one irrational root, because the coefficients are irrational.
A statement like "the roots are real for every real $k
quot; must hold for all $k$, including values that make the leading coefficient zero; check those separately.

Sum and product of roots (Vieta's relations)

If $\alpha,\beta$ are the roots of $ax^2+bx+c=0$, then $a(x-\alpha)(x-\beta)=ax^2+bx+c$ for all $x$. Comparing the coefficients of $x$ and the constant term gives $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. These hold whether the roots are real or complex, and also for repeated roots.

The coefficients thus encode the roots, and many conditions on the roots turn into conditions on the coefficients without solving the equation. Roots equal in magnitude and opposite in sign mean $\alpha+\beta=0$, that is $b=0$. Reciprocal roots ($\alpha\beta=1$) mean $a=c$. One root zero means $c=0$. Both roots positive (real) need $D\ge 0$, $-\frac ba>0$ and $\frac ca>0$.

The distance between the roots comes from $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{D}{a^2}$, so $|\alpha-\beta|=\frac{\sqrt D}{|a|}$.

If the roots are in the ratio $m:n$, write them as $m r$ and $n r$. Then $-\frac ba=(m+n)r$ and $\frac ca=mnr^2$; eliminating $r$ gives the condition $mn\,b^2=(m+n)^2ac$. For instance $x^2-5x+6=0$ has roots $2,3$ in the ratio $2:3$, and indeed $2\cdot3\cdot25=150=25\cdot 6$.

Key points

Sum of roots $=-\frac ba$; product of roots $=\frac ca$.
Opposite roots: $b=0$. Reciprocal roots: $c=a$. One root zero: $c=0$.
Both roots positive: $D\ge0$, $\frac ca>0$, $-\frac ba>0$. Both negative: $D\ge0$, $\frac ca>0$, $\frac ba>0$. Roots of opposite sign: $\frac ca<0$.
Roots in ratio $m:n$: $mn\,b^2=(m+n)^2ac$.
$|\alpha-\beta|=\dfrac{\sqrt{b^2-4ac}}{|a|}$.

Vieta for a quadratic: $\alpha+\beta=-\dfrac ba,\qquad \alpha\beta=\dfrac ca$
Gap between roots: $|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}$
Roots in ratio $m:n$: $mn\,b^2=(m+n)^2\,ac$

Common traps

Vieta's relations give the sum and product of both roots, which are real only when $D\ge0$. A condition derived purely from sum and product (like "both roots positive") also needs $D\ge0$ for the roots to exist as real numbers.
Writing the roots as $-\frac{b}{a}$ and $\frac{c}{a}$ individually is incorrect: those are their sum and product.

Shortcuts

Ratio-of-roots condition. When the roots are in a given ratio $m:n$, skip solving: use $mn\,b^2=(m+n)^2ac$ directly. For roots in ratio $2:1$ this reads $2b^2=9ac$.

Forming an equation from given roots

A monic quadratic with roots $\alpha,\beta$ is $x^2-(\alpha+\beta)x+\alpha\beta=0$. To build an equation whose roots are some expressions in $\alpha,\beta$, compute the sum and the product of the new roots using the known $\alpha+\beta$ and $\alpha\beta$ from the old equation, and plug them into this template.

For simple transformations there is a faster, substitution-based method. If the new roots are $\alpha+h$ (shifted by $h$), replace $x$ by $x-h$ in the old equation. If the new roots are $k\alpha$, replace $x$ by $\frac xk$. If the new roots are $-\alpha$, replace $x$ by $-x$. If the new roots are $\frac1\alpha$ (with $c\neq0$), the equation $ax^2+bx+c=0$ becomes $cx^2+bx+a=0$: the coefficients are reversed.

Example: $x^2-5x+6=0$ has roots $2,3$. The equation with roots $\alpha+2$ is $(x-2)^2-5(x-2)+6=x^2-9x+20=0$, with roots $4,5$. The equation with roots $3\alpha$ is $\left(\frac x3\right)^2-5\cdot\frac x3+6=0$, which on multiplying by $9$ is $x^2-15x+54=0$, with roots $6,9$. The equation with roots $\frac1\alpha,\frac1\beta$ is $6x^2-5x+1=0$, with roots $\frac12,\frac13$.

Key points

Monic equation with roots $\alpha,\beta$: $x^2-(\alpha+\beta)x+\alpha\beta=0$.
Roots $\alpha+h$: substitute $x\to x-h$. Roots $k\alpha$: substitute $x\to x/k$. Roots $-\alpha$: substitute $x\to -x$.
Roots $\frac1\alpha$: reverse the coefficients ($a\leftrightarrow c$).
Roots $\alpha^2,\beta^2$: sum $S^2-2P$, product $P^2$, where $S=\alpha+\beta$, $P=\alpha\beta$.

Equation from roots: $x^2-(\alpha+\beta)x+\alpha\beta=0$
Roots $\alpha^2,\beta^2$: $x^2-\left(S^2-2P\right)x+P^2=0$ (with $S=\alpha+\beta$ and $P=\alpha\beta$ of the original equation)

Common traps

Multiplying an equation by a constant does not change its roots, so "the" equation with given roots is only fixed up to a constant factor. Compare ratios of coefficients, not the coefficients themselves.
The reciprocal-roots rule requires $c\neq0$; if $c=0$ one root is $0$ and has no reciprocal.

Shortcuts

Transform the roots by substitution. To get roots shifted by $h$, scaled by $k$ or negated, substitute $x\to x-h$, $x\to \frac xk$ or $x\to -x$ in the given equation; to get reciprocal roots, swap $a$ and $c$. No sum or product is needed. Example: roots of $x^2-5x+6=0$ raised by $2$ gives $x^2-9x+20=0$.

Symmetric functions of the roots

An expression in $\alpha$ and $\beta$ that stays the same when the two are swapped is symmetric. Every symmetric polynomial can be written using only $S=\alpha+\beta$ and $P=\alpha\beta$, which Vieta supplies, so its value can be found without finding the roots.

The workhorse identities are $\alpha^2+\beta^2=S^2-2P$, $\alpha^3+\beta^3=S^3-3PS$, $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2P^2$ and $(\alpha-\beta)^2=S^2-4P$. For differences, $\alpha^3-\beta^3=(\alpha-\beta)(S^2-P)$. Reciprocal sums: $\frac1\alpha+\frac1\beta=\frac SP$ and $\frac\alpha\beta+\frac\beta\alpha=\frac{S^2-2P}{P}$.

For higher powers, use the fact that each root satisfies the equation: $\alpha^2=S\alpha-P$ and similarly for $\beta$. Multiplying by $\alpha^{n-2}$ and adding the analogue for $\beta$ gives the recurrence $s_n=S\,s_{n-1}-P\,s_{n-2}$ for $s_n=\alpha^n+\beta^n$, starting from $s_0=2$, $s_1=S$. For $x^2-3x+1=0$: $s_2=7,\ s_3=18,\ s_4=47,\ s_5=123$.

Non-symmetric expressions (such as $\alpha^2\beta$ alone) have no value fixed by $S$ and $P$; only the symmetric combination (for instance $\alpha^2\beta+\alpha\beta^2=PS$) does.

Key points

$\alpha^2+\beta^2=S^2-2P$ and $\alpha^3+\beta^3=S^3-3SP$.
$(\alpha-\beta)^2=S^2-4P$, so $\alpha-\beta=\pm\sqrt{S^2-4P}$ (sign depends on the labelling).
$\frac1\alpha+\frac1\beta=\frac SP$.
$s_n=S\,s_{n-1}-P\,s_{n-2}$ for power sums $s_n=\alpha^n+\beta^n$.
$\alpha^2\beta+\alpha\beta^2=SP$.

Square and cube sums: $\alpha^2+\beta^2=S^2-2P,\qquad \alpha^3+\beta^3=S^3-3SP$
Fourth power sum: $\alpha^4+\beta^4=(S^2-2P)^2-2P^2$
Power-sum recurrence: $s_n=S\,s_{n-1}-P\,s_{n-2}$ ($s_n=\alpha^n+\beta^n$, $s_0=2$, $s_1=S$)
Difference of cubes: $\alpha^3-\beta^3=(\alpha-\beta)\left(S^2-P\right)$

Common traps

$\alpha-\beta$ alone is not symmetric; only $(\alpha-\beta)^2$ is. Its value is fixed but its sign is not, which matters when an answer asks for $\alpha-\beta$ with a given ordering.
The identity $\alpha^2+\beta^2=S^2-2P$ holds for complex roots too, but the result may be negative when the roots are non-real.

Shortcuts

Power sums by recurrence. Instead of expanding each power, use $s_n=S\,s_{n-1}-P\,s_{n-2}$ with $s_0=2,\ s_1=S$. For $x^2-3x+1=0$, $S=3$, $P=1$, so $s_2=7,\ s_3=18,\ s_4=47,\ s_5=123$, each from the two before it.

Common roots of two quadratics

Two quadratics $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ can share one root or both. To find a common root, eliminate $x^2$: multiply the first by $a_2$, the second by $a_1$, and subtract. The result is linear in $x$ and gives the only possible common root $x=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$ (when $a_1b_2\neq a_2b_1$).

Substituting that value back into either equation gives the condition for exactly one common root: $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$. This condition is the vanishing of the resultant of the two quadratics.

Both roots are common exactly when the equations are proportional: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$. Two further shortcuts help. If the coefficients are real and the equations share one non-real root, they share its conjugate as well, so both roots are common. If the coefficients are rational and they share one irrational root $p+\sqrt q$, they share the conjugate $p-\sqrt q$ as well.

A standard example: $x^2+px+q=0$ and $x^2+qx+p=0$. Subtracting gives $(p-q)(x-1)=0$. Either $p=q$ (the equations coincide) or the common root is $x=1$, which forces $1+p+q=0$. The resultant factorises as $(p-q)^2(p+q+1)$, which confirms both cases.

Key points

Subtracting the (scaled) equations gives a linear equation, whose solution is the only possible common root.
One common root: $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$.
Both roots common: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.
Real coefficients and one common non-real root imply both roots are common (conjugate pair).

Common root value: $x=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$
Condition for one common root: $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$

Common traps

The proportionality condition is for both roots common. The one-root condition also holds when both are common (in that case it holds trivially as $0=0$), so on its own it does not guarantee that exactly one root is shared.
After finding a parameter value from the one-root condition, check the case where the two equations become identical: a separate value of the parameter can make all roots common.

Shortcuts

Eliminate x squared when the leading coefficients match. If both quadratics have equal leading coefficients, subtracting them leaves a linear equation, which names the only possible common root; then test it. For $x^2+kx+2$ and $x^2+2x+k$, subtracting gives $(k-2)(x-1)=0$, so the only candidate is $x=1$ (or the equations coincide when $k=2$). Substituting $x=1$ gives $1+k+2=0$, so $k=-3$; for any other $k\neq2$ there is no common root.

Graph of a quadratic and the sign of the expression

The graph of $y=ax^2+bx+c$ is a parabola. It opens upward for $a>0$ and downward for $a<0$. Its axis of symmetry is $x=-\frac{b}{2a}$, its vertex is $\left(-\frac{b}{2a},\,-\frac{D}{4a}\right)$, and it meets the $y$-axis at $(0,c)$. The $x$-intercepts are the real roots, so the sign of $D$ decides whether the parabola cuts, touches or misses the $x$-axis.

The sign of $f(x)=ax^2+bx+c$ follows from the picture. Between the roots (when $D>0$), $f$ has the sign opposite to $a$; outside the roots it has the same sign as $a$. If $D<0$ the parabola never reaches the axis, so $f(x)$ has the sign of $a$ for every real $x$. If $D=0$, $f(x)$ has the sign of $a$ everywhere except at the repeated root where it is $0$.

This produces the condition for a quadratic to be always positive: $a>0$ and $D<0$. It is always non-negative: $a>0$ and $D\le0$. For always negative: $a<0$ and $D<0$. For example $x^2+2(k+1)x+9k-5>0$ for all real $x$ requires $4(k+1)^2-4(9k-5)<0$, which simplifies to $k^2-7k+6<0$, so $1<k<6$.

The quantity $a\,f(k)$ is a compact sign test for a point $k$: since $a\,f(k)=\left(ak+\frac b2\right)^2-\frac D4$, a negative value means $D>0$ and $k$ lies strictly between the roots; a positive value means $k$ lies outside the closed interval between the roots (when $D>0$).

Key points

Vertex $\left(-\frac{b}{2a},-\frac{D}{4a}\right)$; axis $x=-\frac{b}{2a}$; $y$-intercept $c$.
$D>0$: graph cuts the $x$-axis twice; $D=0$: touches once; $D<0$: no contact.
Sign: outside the roots same as $a$, between the roots opposite to $a$.
$f(x)>0$ for all real $x$ iff $a>0$ and $D<0$; $f(x)<0$ for all $x$ iff $a<0$ and $D<0$.
$a\,f(k)<0$ iff $k$ lies strictly between the roots.

Vertex: $\left(-\dfrac{b}{2a},\ -\dfrac{b^2-4ac}{4a}\right)$
Sign test: $a\,f(k)=\left(ak+\dfrac b2\right)^2-\dfrac{D}{4}$

Common traps

"Always positive" with a parameter in the leading coefficient: $a>0$ is needed, but also check the value that makes $a=0$, where the expression becomes linear or constant and may or may not qualify.
For a non-strict inequality $f(x)\ge0$ for all $x$, $D=0$ is allowed; for $f(x)>0$ it is not.

Location of roots relative to a number or an interval

Many problems ask where the roots of $f(x)=ax^2+bx+c$ lie relative to given numbers, with a parameter inside the coefficients. Conditions are built from four ingredients: whether the roots are real ($D\ge0$), the position of the axis $x=-\frac{b}{2a}$, the sign of $a\,f(k)$ at the reference point $k$, and sometimes the sign of $f$ at both ends of an interval. Multiplying $f(k)$ by $a$ removes the need to split into cases for $a>0$ and $a<0$.

The standard cases, with $\alpha\le\beta$ the roots: both roots greater than $k$: $D\ge0$, $-\frac b{2a}>k$, $a\,f(k)>0$. Both roots less than $k$: $D\ge0$, $-\frac b{2a}<k$, $a\,f(k)>0$. The number $k$ strictly between the roots: $a\,f(k)<0$ (and nothing else). Both roots inside $(k_1,k_2)$: $D\ge0$, $k_1<-\frac b{2a}<k_2$, $a\,f(k_1)>0$, $a\,f(k_2)>0$. Exactly one root inside $(k_1,k_2)$: $f(k_1)f(k_2)<0$. Both $k_1,k_2$ between the roots, so that $\alpha<k_1<k_2<\beta$: $a\,f(k_1)<0$ and $a\,f(k_2)<0$.

Example: find $m$ such that both roots of $x^2-2mx+m+2=0$ exceed $1$. Here $D=4m^2-4m-8\ge0$ gives $m\le-1$ or $m\ge2$; the axis $x=m>1$; and $f(1)=3-m>0$ gives $m<3$. Combining, $2\le m<3$.

A single sign condition at one point also tells whether the roots lie on opposite sides of $0$: $f(0)=c$, so $a\,f(0)=ac<0$ means $0$ lies between the roots.

Key points

Both roots
gt;k$: $D\ge0,\ -\frac b{2a}>k,\ a f(k)>0$.
Both roots
lt;k$: $D\ge0,\ -\frac b{2a}<k,\ a f(k)>0$.
$k$ between the roots: $a f(k)<0$.
Both roots in $(k_1,k_2)$: $D\ge0$, $k_1<-\frac b{2a}<k_2$, $a f(k_1)>0$, $a f(k_2)>0$.
Exactly one root in $(k_1,k_2)$: $f(k_1)f(k_2)<0$ is sufficient; it is necessary only when neither $k_1$ nor $k_2$ is itself a root (endpoint roots and a double root need separate checks).
Roots on opposite sides of $0$: $ac<0$.

Both roots greater than $k$: $D\ge0,\quad -\dfrac{b}{2a}>k,\quad a\,f(k)>0$
$k$ between the roots: $a\,f(k)<0$
Both roots in $(k_1,k_2)$: $D\ge0,\ \ k_1<-\dfrac{b}{2a}<k_2,\ \ a\,f(k_1)>0,\ \ a\,f(k_2)>0$

Common traps

The condition $f(k_1)f(k_2)<0$ is sufficient for exactly one root in the open interval, but not necessary: if $f(k_1)=0$ or $f(k_2)=0$ the product is $0$, and the other root has to be tested separately.
Dropping the axis condition gives a wrong answer: $D\ge0$ and $a\,f(k)>0$ alone cannot tell "both roots greater than $k
quot; from "both roots less than $k
quot;.
Whether to use $D\ge0$ or $D>0$ depends on whether a repeated root counts as "two roots" in the problem statement.

Shortcuts

One test for a point between the roots. To check that a number $k$ lies strictly between the two roots, the single condition $a\,f(k)<0$ is enough, because $D=(2ak+b)^2-4\,a\,f(k)$ is then positive automatically. Example: for $x^2-5x+6$ and $k=2.5$, $f(k)=-0.25<0$, and the roots are $2$ and $3$.

Range, maximum and minimum of a quadratic expression

For $f(x)=ax^2+bx+c$ the vertex gives the extreme value. If $a>0$ the minimum is $-\frac{D}{4a}$, attained at $x=-\frac b{2a}$, and the range is $\left[-\frac D{4a},\infty\right)$. If $a<0$ the maximum is $-\frac D{4a}$ at the same $x$, and the range is $\left(-\infty,-\frac D{4a}\right]$. Example: $x^2-6x+13$ has vertex $(3,4)$, so its minimum is $4$.

On a closed interval $[p,q]$ the extreme values occur at the endpoints or at the vertex if it lies inside the interval. For $x^2-6x+13$ on $[4,6]$ the vertex $x=3$ is outside, so $f$ is increasing there: the minimum is $f(4)=5$ and the maximum is $f(6)=13$.

To find the range of a rational expression such as $y=\dfrac{x^2+x+1}{x^2-x+1}$, treat $y$ as a fixed number and ask for which $y$ the equation has a real $x$. Cross-multiplying gives $(y-1)x^2-(y+1)x+(y-1)=0$. For real $x$ the discriminant must be non-negative: $(y+1)^2-4(y-1)^2\ge0$, which factorises as $-(y-3)(3y-1)\ge0$, so $\frac13\le y\le3$. The value $y=1$ (where the coefficient of $x^2$ vanishes) is attained at $x=0$, and the denominator never vanishes since its discriminant is negative.

Related facts: for two numbers with a fixed sum $s$, the product $x(s-x)$ is largest, $s^2/4$, when $x=s/2$; for positive numbers with a fixed product the sum is smallest when they are equal.

Key points

$a>0$: minimum $-\frac{D}{4a}$ at $x=-\frac b{2a}$. $a<0$: maximum $-\frac D{4a}$ there.
On $[p,q]$: compare $f(p)$, $f(q)$ and, if $p\le-\frac b{2a}\le q$, the vertex value.
Range of a quadratic-over-quadratic: write $y\cdot(\text{denominator})=\text{numerator}$ and require $D_x\ge0$.
With fixed sum $s$, the product $x(s-x)$ has maximum $\frac{s^2}{4}$.

Extreme value: $f_{\text{ext}}=-\dfrac{D}{4a}=c-\dfrac{b^2}{4a}$ (minimum if $a>0$, maximum if $a<0$, at $x=-\dfrac{b}{2a}$)

Common traps

The vertex value is the extreme value over all real $x$ only; on a restricted interval the vertex may lie outside it and the answer comes from an endpoint.
In the discriminant method for rational functions, the coefficient of $x^2$ may vanish for one special $y$; check that value separately because the equation becomes linear there. Also exclude any $y$ that arises only from a common factor of numerator and denominator (a removable point).

Quadratic and rational inequalities, the wavy curve method

To solve $f(x)>0$ or $f(x)<0$ for a quadratic, factorise and use the sign picture: for $a>0$ and roots $\alpha<\beta$, $f(x)<0$ on $(\alpha,\beta)$ and $f(x)>0$ outside $[\alpha,\beta]$. Example: $x^2-5x+6<0$ gives $2<x<3$, and $x^2>4$ gives $x<-2$ or $x>2$. For $-x^2+x+6\ge0$, factorise as $-(x-3)(x+2)\ge0$ and conclude $-2\le x\le3$.

For products and quotients of several factors the wavy curve (sign scheme) method scales to any number of factors. Write the expression as a product of powers of linear factors $(x-r)^m$ with the coefficient of $x$ positive in each factor. Mark the critical points on a number line in increasing order. Start with the sign $+$ to the right of the largest critical point (with an overall positive constant). Moving left, the sign flips when passing a root of odd multiplicity and stays the same when passing a root of even multiplicity.

Take $\dfrac{(x+2)^2(x-1)(x-4)}{x-3}\le0$. Critical points: $-2$ (even power), $1$ and $4$ (numerator, included for $\le$), $3$ (denominator, always excluded). To the right of $4$ the sign is $+$; in $(3,4)$ it is $-$; in $(1,3)$ it is $+$; in $(-2,1)$ it is $-$; to the left of $-2$ the sign stays $-$ because $-2$ has an even power. So the solution set is $(-\infty,1]\cup(3,4]$, where $x=-2$ belongs to it because the expression equals $0$ there.

Points where the expression equals $0$ because of an even-power factor belong to the solution of a non-strict inequality even though the sign does not change there. Never multiply a rational inequality by the denominator unless its sign is known; rewrite as a single fraction compared with $0$ instead.

Key points

Factorise fully; put every factor in the form $(x-r)$.
Sign flips at odd-multiplicity roots only; even-multiplicity roots touch the axis and keep the sign.
Roots of the denominator are always excluded; roots of the numerator are included for $\le$ or $\ge$.
$x^2>c^2$ means $x<-|c|$ or $x>|c|$; $x^2<c^2$ means $-|c|<x<|c|$.

Quadratic sign pattern ($a>0$, roots $\alpha<\beta$): $(x-\alpha)(x-\beta)<0\iff \alpha<x<\beta$

Common traps

$x^2>4$ is not the same as $x>\pm2$: it is $x<-2$ or $x>2$ (two separate intervals).
A factor like $(2-x)$ has a negative coefficient of $x$. Rewrite it as $-(x-2)$; each such factor contributes one factor of $-1$, so multiply the whole inequality by $-1$ once per rewritten factor, reversing the inequality each time. With two negated factors the two reversals cancel. Example: $(2-x)(x+1)\ge0$ becomes $(x-2)(x+1)\le0$, giving $-1\le x\le2$.
Cancelling a common factor between numerator and denominator removes a point that is still excluded from the domain.

Shortcuts

Starting sign for the wavy curve. After writing every factor as $(x-r)$, the interval right of the largest critical point is always positive, so begin there and alternate going left, skipping the flip at even powers. For $(x-1)(x-3)\ge 0$ the signs are $+,-,+$ from the left: $x\le1$ or $x\ge3$.

Drop a factor that is always positive. A quadratic factor with $a>0$ and $D<0$ is positive for every real $x$, so it can be deleted from an inequality without changing the solution. Example: $(x^2+x+1)(x-1)(x-3)\ge0$ has the same solution as $(x-1)(x-3)\ge0$, namely $x\le1$ or $x\ge3$, because $x^2+x+1$ has $D=-3<0$.

Equations reducible to quadratics

Many equations become quadratic after a substitution. If an equation contains only $x^{2n}$ and $x^n$ (for example $x^4-5x^2+4=0$), put $t=x^n$ to get $t^2-5t+4=0$, so $t=1,4$ and $x=\pm1,\pm2$. If a repeated expression appears, such as $x^2+3x$ in $(x^2+3x)^2-2(x^2+3x)-8=0$, set $t=x^2+3x$: then $t^2-2t-8=0$ gives $t=4$ or $-2$, and $x^2+3x-4=0$ gives $x=1,-4$ while $x^2+3x+2=0$ gives $x=-1,-2$.

Reciprocal-type equations: a polynomial whose coefficients read the same forwards and backwards, like $2x^4-3x^3-x^2-3x+2=0$, is divided by $x^2$ and written in terms of $t=x+\frac1x$, using $x^2+\frac1{x^2}=t^2-2$. This gives $2(t^2-2)-3t-1=0$, that is $2t^2-3t-5=0$, so $t=\frac52$ or $t=-1$. Then $x+\frac1x=\frac52$ gives $x=2,\frac12$, while $x+\frac1x=-1$ has no real solution (it gives $x^2+x+1=0$).

Exponential equations: $4^x-3\cdot2^x+2=0$ becomes $t^2-3t+2=0$ with $t=2^x>0$, so $t=1,2$ and $x=0,1$. The substitution variable must respect the range of the original function; $2^x$ is always positive, so negative values of $t$ must be rejected.

Radical equations: $\sqrt{2x+3}=x$ is squared to $x^2-2x-3=0$, so $x=3$ or $x=-1$. Squaring can introduce extra roots: $x=-1$ gives $\sqrt1=1\neq-1$, so only $x=3$ works. Logarithmic equations likewise need the domain check after solving the quadratic.

Key points

Substitute $t=x^n$ for equations in $x^{2n}$ and $x^n$.
Repeated expression: let it be $t$, solve, then solve each $x$-equation.
Reciprocal equations: divide by $x^2$, use $t=x+\frac1x$, $x^2+\frac1{x^2}=t^2-2$.
Always check the range of the substitute ($t=2^x>0$, $t=x+\frac1x$ has $|t|\ge2$ for real $x$) and verify radical or logarithmic solutions in the original.

Reciprocal substitution: $x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2$

Common traps

For real $x$, $t=x+\frac1x$ satisfies $t\ge2$ or $t\le-2$; a value of $t$ in $(-2,2)$ gives only non-real $x$.
Solving for $t$ is half the work: each admissible $t$ then needs its own equation in $x$, and the number of real roots is the total over all of them.
After squaring a radical equation, check every candidate in the original, since the right side must be non-negative.

Complex roots of quadratics

When $D<0$ and the coefficients are real, $\sqrt{D}=i\sqrt{-D}$ and the roots are $\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$. They are complex conjugates, so their sum $-\frac ba$ and product $\frac ca$ are real. Example: $x^2-2x+5=0$ has roots $1\pm2i$; the real quadratic with root $3-2i$ is $x^2-6x+13=0$.

The conjugate-pair rule needs real coefficients. If a coefficient is complex, one root need not force its conjugate. For instance $x^2-(1+i)x+i=0$ has roots $1$ and $i$, and its discriminant is $(1+i)^2-4i=-2i$, so the formula still works with a complex square root, but the roots are not conjugates.

The nonreal cube roots of unity, $\omega=\frac{-1+i\sqrt3}{2}$ and $\omega^2=\bar\omega$, are the two roots of $x^2+x+1=0$, with $\omega^3=1$ and $1+\omega+\omega^2=0$.

To find a real root of an equation with complex coefficients, substitute a real $x$, separate real and imaginary parts, and solve both parts together. In $x^2-(2+i)x+(1+i)=0$ the imaginary part is $-x+1=0$, so $x=1$, and the real part $x^2-2x+1=0$ is also satisfied at $x=1$. The other root is $1+i$ (the discriminant is $(2+i)^2-4(1+i)=-1$).

Key points

Real coefficients and $D<0$: roots $\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$, a conjugate pair.
Real quadratic with root $p+iq$ ($q\ne0$): $x^2-2px+(p^2+q^2)=0$.
Cube roots of unity: $1,\omega,\omega^2$ with $1+\omega+\omega^2=0$.
The formula works for complex coefficients; the conjugate-pair property does not.

Real quadratic with root $p+iq$: $x^2-2px+\left(p^2+q^2\right)=0$
Cube roots of unity: $\omega^2+\omega+1=0,\qquad \omega^3=1$

Common traps

Inequalities such as $D\ge0$ describe reality of roots only for real coefficients; with complex coefficients, "real roots" has to be tested by splitting into real and imaginary parts.
The two roots $\alpha,\beta$ of a real quadratic with $D<0$ cannot be ordered, so statements like $\alpha<\beta$ are meaningless; only $|\alpha|$, real part and imaginary part are comparable.

Quadratics with a modulus

Since $|x|^2=x^2$, an equation like $x^2-3|x|+2=0$ is a quadratic in $t=|x|$: $t^2-3t+2=0$ gives $t=1$ or $2$. As $t=|x|\ge0$ only non-negative values of $t$ are allowed, and every positive $t$ gives two values of $x$ ($\pm t$), while $t=0$ gives one. So $x=\pm1,\pm2$, four real roots. In $x^2-|x|-6=0$, $t^2-t-6=0$ gives $t=3$ or $t=-2$; the negative value is rejected, leaving $x=\pm3$.

For $|f(x)|=k$ with $k>0$ the equation splits into $f(x)=k$ or $f(x)=-k$. Example: $|x^2-4x|=3$ gives $x^2-4x-3=0$ with $x=2\pm\sqrt7$, and $x^2-4x+3=0$ with $x=1,3$: four roots in all, and each satisfies $|x^2-4x|=3$.

Inequalities with a modulus follow the same idea: $|x-a|<b$ ($b>0$) means $a-b<x<a+b$, and $|x-a|>b$ means $x<a-b$ or $x>a+b$. For instance $|x-2|<3$ gives $-1<x<5$.

If the question is how many real roots $ax^2+b|x|+c=0$ has, find the roots of $at^2+bt+c=0$ and count: each positive root gives two real $x$, a zero root gives one, and negative roots give none.

Key points

$|x|^2=x^2$, so $x^2$ and $|x|^2$ are interchangeable.
In $at^2+bt+c=0$ with $t=|x|$: positive roots count twice, zero once, negative roots not at all.
$|f|=k$ ($k\ge0$): $f=\pm k$. $|f|=|g|$: $f=g$ or $f=-g$.
$|x-a|<b\iff a-b<x<a+b$ for $b>0$.

Modulus inequality: $|x-a|<b\iff a-b<x<a+b\quad (b>0)$

Common traps

A root of the $t$-equation that is negative must be discarded, since $t=|x|$ cannot be negative.
For $|f(x)|=g(x)$ the right side must satisfy $g(x)\ge0$; the solutions of $f=g$ and $f=-g$ are valid only where $g\ge0$.
The equation $|x|=t$ has two solutions only when $t>0$.

Cubic and biquadratic equations through Vieta's relations

For $ax^3+bx^2+cx+d=0$ with roots $\alpha,\beta,\gamma$: $\alpha+\beta+\gamma=-\frac ba$, $\alpha\beta+\beta\gamma+\gamma\alpha=\frac ca$, $\alpha\beta\gamma=-\frac da$. For $ax^4+bx^3+cx^2+dx+e=0$ with roots $\alpha,\beta,\gamma,\delta$: the sums of the roots taken one, two, three and four at a time are $-\frac ba,\ \frac ca,\ -\frac da,\ \frac ea$. The signs alternate, starting with a minus. With $\sigma_1,\sigma_2,\sigma_3$ for the cubic, the equation from its roots is $x^3-\sigma_1x^2+\sigma_2x-\sigma_3=0$.

Symmetric functions follow the same pattern as in the quadratic case: $\alpha^2+\beta^2+\gamma^2=\sigma_1^2-2\sigma_2$, $\alpha^3+\beta^3+\gamma^3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$, and $\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\sigma_2}{\sigma_3}$. Also $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=\sigma_1\left(\alpha^2+\beta^2+\gamma^2-\sigma_2\right)$. Example: $x^3-2x^2+3x-4=0$ has $\sigma_1=2$, $\sigma_2=3$, $\sigma_3=4$, so $\sum\alpha^2=4-6=-2$, $\sum\alpha^3=8-18+12=2$ and $\sum\frac1\alpha=\frac34$. The negative value of $\sum\alpha^2$ shows that some roots are non-real: indeed only one real root exists, since a real cubic has either one or three real roots, counting multiplicity.

Reciprocal roots again mean reversing coefficients: the equation with roots $\frac1\alpha,\frac1\beta,\frac1\gamma$ for the example is $-4x^3+3x^2-2x+1=0$.

If integer coefficients are given and a rational root $\frac pq$ (in lowest terms) exists, then $p$ divides the constant term and $q$ divides the leading coefficient. For $2x^3-3x^2-3x+2=0$ the candidates lead to the roots $2,\ \frac12,\ -1$. When the roots are said to be in a progression, parametrise them to match: in AP use $m-d,m,m+d$; in GP use $\frac mq,m,mq$.

Key points

Cubic: $\sigma_1=-\frac ba$, $\sigma_2=\frac ca$, $\sigma_3=-\frac da$.
Quartic: sums of roots by 1, 2, 3, 4 at a time are $-\frac ba,\ \frac ca,\ -\frac da,\ \frac ea$.
Roots in AP: the middle root is $-\frac{b}{3a}$. Roots in GP: the middle root cubed is $-\frac da$.
A real cubic always has at least one real root.
Rational root test: $\frac pq$ with $p\mid d$ and $q\mid a$.

Cubic from roots: $x^3-\sigma_1x^2+\sigma_2x-\sigma_3=0$
Cubic power sums: $\sum\alpha^2=\sigma_1^2-2\sigma_2,\qquad \sum\alpha^3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$
Sum of reciprocals: $\sum\dfrac1\alpha=\dfrac{\sigma_2}{\sigma_3}$

Common traps

Vieta for the cubic uses alternating signs: the sum of roots is $-\frac ba$, the product is $-\frac da$ (negative of the constant over $a$), while the pairwise sum is $+\frac ca$.
A sum of squares of real numbers is non-negative, so $\sum\alpha^2<0$ is possible only when some roots are non-real, as in the example.
For a quartic, "sum of products taken three at a time" is $-\frac da$ (the coefficient of $x$), not the coefficient of $x^2$.

Shortcuts

Roots in arithmetic progression. If the three roots of $ax^3+bx^2+cx+d=0$ are in AP, the middle root is $-\frac b{3a}$ (the sum of the roots is $3\times$ the middle one). Substitute it into the equation to get a condition. Example: roots $1,2,3$ of $x^3-6x^2+11x-6$ have middle root $\frac{6}{3}=2$.

Roots in geometric progression. If the three roots are in GP, the middle root $\beta$ satisfies $\beta^3=-\frac da$, because $\frac\beta r\cdot\beta\cdot\beta r=\beta^3$. Example: roots $1,2,4$ of $x^3-7x^2+14x-8$ give $\beta^3=8$, so $\beta=2$.

Worked examples

Both roots larger than 1

Find all real $m$ for which both roots of $x^2-2mx+m+2=0$ are real and greater than $1$.

List the conditions. With $f(x)=x^2-2mx+m+2$ and $a=1>0$, both roots greater than $1$ means $D\ge0$, axis $x=m>1$, and $f(1)>0$.
Discriminant. $D=4m^2-4(m+2)=4(m^2-m-2)=4(m-2)(m+1)\ge0$, so $m\le-1$ or $m\ge2$.
Axis and value at 1. The axis gives $m>1$. Also $f(1)=1-2m+m+2=3-m>0$, so $m<3$.
Intersect. Combining $m\le-1$ or $m\ge2$, with $m>1$ and $m<3$ leaves $2\le m<3$.
Spot check. At $m=2$: $x^2-4x+4=0$ has the double root $2>1$. At $m=3$: $x^2-6x+5=0$ has roots $1,5$, and $1$ is not greater than $1$, so $m=3$ is excluded.

Answer: $2\le m<3$

A parameter forcing a common root

Find all real $k$ for which $x^2+kx+2=0$ and $x^2+2x+k=0$ have at least one common root.

Subtract the equations. A common root satisfies both, so it satisfies their difference: $(k-2)x+(2-k)=0$, that is $(k-2)(x-1)=0$.
Case k not equal to 2. Then $x=1$ is the common root. Substituting in the first equation, $1+k+2=0$, so $k=-3$.
Check k = -3. The equations become $x^2-3x+2=0$ with roots $1,2$ and $x^2+2x-3=0$ with roots $1,-3$. They share exactly the root $1$.
Case k = 2. Both equations are $x^2+2x+2=0$, identical, with common (complex) roots $-1\pm i$. So $k=2$ also works, with both roots common.
Cross-check. The resultant of the two quadratics factorises as $(k-2)^2(k+3)$, vanishing exactly at $k=2$ and $k=-3$.

Answer: $k=-3$ (one common root, $x=1$) or $k=2$ (the equations coincide)

Range of a ratio of quadratics

Find the range of $y=\dfrac{x^2+x+1}{x^2-x+1}$ for real $x$.

Clear the denominator. The denominator has discriminant $1-4<0$, so it never vanishes. Write $y(x^2-x+1)=x^2+x+1$, which rearranges to $(y-1)x^2-(y+1)x+(y-1)=0$.
Real x needs D at least 0. The discriminant in $x$ is $(y+1)^2-4(y-1)^2=-(y-3)(3y-1)$. It must be non-negative.
Solve for y. $(y-3)(3y-1)\le0$ gives $\frac13\le y\le3$.
The special value y = 1. At $y=1$ the equation becomes linear: $-2x=0$, so $x=0$ and $y=1$ is attained (inside the interval anyway).
Endpoints are attained. At $x=1$, $y=\frac{3}{1}=3$. At $x=-1$, $y=\frac{1}{3}$.

Answer: Range $\left[\frac13,\,3\right]$

Symmetric expressions without the roots

If $\alpha,\beta$ are the roots of $x^2-5x+3=0$, find $\alpha^2+\beta^2$, $\alpha^4+\beta^4$, $\frac\alpha\beta+\frac\beta\alpha$, $(\alpha-\beta)^2$, and the equation with roots $\frac\alpha\beta$ and $\frac\beta\alpha$.

Read S and P. $S=\alpha+\beta=5$ and $P=\alpha\beta=3$.
Squares. $\alpha^2+\beta^2=S^2-2P=25-6=19$. Also $(\alpha-\beta)^2=S^2-4P=25-12=13$.
Fourth powers. $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2P^2=361-18=343$.
The ratio sum. $\frac\alpha\beta+\frac\beta\alpha=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{19}{3}$.
New equation. The product of $\frac\alpha\beta$ and $\frac\beta\alpha$ is $1$, and their sum is $\frac{19}3$. So the equation is $x^2-\frac{19}{3}x+1=0$, or $3x^2-19x+3=0$.

Answer: $19,\ 343,\ \frac{19}{3},\ 13$, and $3x^2-19x+3=0$

A rational inequality with an even power

Solve $\dfrac{(x+2)^2(x-1)(x-4)}{x-3}\le0$.

Critical points. Numerator zeros: $-2$ (even power), $1$, $4$. Denominator zero: $3$ (excluded).
Rightmost interval. All factors are written as $(x-r)$, so for $x>4$ the expression is positive.
Move left. Crossing $4$ (odd) flips to negative on $(3,4)$. Crossing $3$ (odd, in the denominator) flips to positive on $(1,3)$. Crossing $1$ flips to negative on $(-2,1)$. Crossing $-2$ (even) does not flip, so $(-\infty,-2)$ is negative too.
Select the intervals. We want $\le0$: the negative intervals $(3,4)$, $(-2,1)$ and $(-\infty,-2)$, together with the zeros $x=1$, $x=4$ and $x=-2$. The point $x=3$ is excluded.
Join them. $(-\infty,-2)\cup\{-2\}\cup(-2,1)\cup\{1\}$ is $(-\infty,1]$. Together with $(3,4]$ this is the answer.

Answer: $x\in(-\infty,1]\cup(3,4]$

Power sums of a cubic's roots

If $\alpha,\beta,\gamma$ are the roots of $x^3-2x^2+3x-4=0$, find $\sum\alpha^2$, $\sum\alpha^3$, $\sum\frac1\alpha$, and the cubic whose roots are the reciprocals.

Read the symmetric sums. $\sigma_1=2$, $\sigma_2=3$, $\sigma_3=4$ (note $\sigma_3=-\frac da=4$).
Squares. $\sum\alpha^2=\sigma_1^2-2\sigma_2=4-6=-2$. A negative sum of squares signals non-real roots.
Cubes. $\sum\alpha^3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3=8-18+12=2$.
Reciprocals. $\sum\frac1\alpha=\frac{\sigma_2}{\sigma_3}=\frac34$. The cubic with reciprocal roots reverses the coefficients: $-4x^3+3x^2-2x+1=0$, that is $4x^3-3x^2+2x-1=0$.

Answer: $-2,\ 2,\ \frac34$, and $4x^3-3x^2+2x-1=0$

A modulus equation with four roots

Solve $|x^2-4x|=3$ and count the real roots.

Split by sign. $|x^2-4x|=3$ means $x^2-4x=3$ or $x^2-4x=-3$.
First branch. $x^2-4x-3=0$ gives $x=\frac{4\pm\sqrt{16+12}}{2}=2\pm\sqrt7$.
Second branch. $x^2-4x+3=0$ gives $(x-1)(x-3)=0$, so $x=1$ or $3$.
Verify. For $x=1$: $|1-4|=3$. For $x=3$: $|9-12|=3$. For $x=2\pm\sqrt7$: $x^2-4x=(x-2)^2-4=7-4=3$, so the modulus is $3$.

Answer: $x=1,\ 3,\ 2+\sqrt7,\ 2-\sqrt7$: four real roots

A reciprocal quartic

Solve $2x^4-3x^3-x^2-3x+2=0$ over the reals.

Divide by x squared. Since $x=0$ is not a root, divide by $x^2$: $2\left(x^2+\frac1{x^2}\right)-3\left(x+\frac1x\right)-1=0$.
Substitute. With $t=x+\frac1x$ we have $x^2+\frac1{x^2}=t^2-2$, so $2(t^2-2)-3t-1=0$, that is $2t^2-3t-5=0$, giving $t=\frac52$ or $t=-1$.
Solve for x. $x+\frac1x=\frac52$ gives $2x^2-5x+2=0$, so $x=2$ or $\frac12$. $x+\frac1x=-1$ gives $x^2+x+1=0$, which has no real roots (indeed $|t|<2$).
Check. For $x=2$: $32-24-4-6+2=0$.

Answer: $x=2$ or $x=\frac12$ (the other two roots are non-real)

A quadratic that is positive for every x

Find all $k$ for which $x^2+2(k+1)x+9k-5>0$ for every real $x$.

Translate. The leading coefficient $1$ is positive, so the condition is $D<0$.
Discriminant. $D=4(k+1)^2-4(9k-5)=4\left(k^2-7k+6\right)=4(k-1)(k-6)$.
Solve. $(k-1)(k-6)<0$ gives $1<k<6$.
Boundary check. At $k=1$ or $k=6$, $D=0$ and the expression touches zero at one point, so it is not strictly positive: both endpoints are excluded.

Answer: $1<k<6$

Figure: Three upward-opening parabolas side by side: one cutting the x-axis at two points labelled alpha and beta (D positive), one touching the axis at a single point (D zero), one lying entirely above the axis (D negative).

Figure: Three parabolas with dashed vertical lines marking reference numbers: both roots to the right of k, k lying between the two roots, and both roots between k1 and k2.

Figure: A number line with points -2, 1, 3 and 4 and a wavy curve above and below it showing the sign pattern plus, minus, plus, minus, minus from right to left; the curve touches the line at -2 without crossing and point 3 is a hollow circle.

Figure: The parabola y = x squared minus 6x plus 13 with its vertex at (3, 4) marked as the minimum and the points (4, 5) and (6, 13) marked as the minimum and maximum on the interval from 4 to 6.

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