Quadratics

A quadratic is polynomial of degree 2, ie x2x^2.

Questions rarely ask about quadratics by itself. Rather, it often appears as part of an equation, an inequality, or an integral, or as a step of finding an integral.

Contents

Notations

standard form

f(x)=ax2+bx+cf(x) = ax^2 + bx + c
zeros:   x=−b±b2−4ac2a    (quadratic formula)\text{zeros: }\; x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \;\;(\text{quadratic formula})
vertex:   (−b2a,f(−b2a))\text{vertex: }\; \left(-\frac {b}{2a}, f\left(-\frac {b}{2a}\right)\right)

factored form

f(x)=a(x−x1)(x−x2)f(x) = a(x - x_1)(x - x_2)
zeros:   x1,x2\text{zeros: }\; x_1, x_2
vertex:   (x1+x22,f(x1+x22))\text{vertex: }\; \left(\frac{x_1+x_2}{2}, f\left(\frac{x_1+x_2}{2}\right)\right)

vertex form

f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k
zeros:   h±−ka\text{zeros: }\; h \pm \sqrt{-\frac{k}{a}}
vertex:   (h,k)\text{vertex: }\; (h, k)

Recognize functions transformations in action.

Converting between notations

To convert to factored or vertex forms, calculate the zeros or vertex from the current form and rewrite in the desired form. If converting to standard form, expand the quadratic.

It is rather common to be (implicitly) asked to convert between standard form. In HL, this is especially needed to invoke the integration rule of

∫1x2+a2 dx=1aarctan⁡xa+C\int \frac{1}{x^2 + a^2}\d x = \frac 1a \arctan \frac xa + C

along with a substitution for xx.

complete the square

Completing the square allows identification of the vertex and axis of symmetry. There are two methods.

The first method involves

  1. factoring out the leading coefficient from first two terms;
  2. adding and subtracting the square of half of the second term in the bracket;
  3. taking the subtracted term out of the bracket, and
  4. Using (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Example: Convert 3x2+4x+53x^2 + 4x + 5 to vertex form.

p(x)=3x2+4x+5=3(x2+43x)+5=3(x2+2⋅23x+(23)2−(23)2)+5=3(x2+2⋅23x+(23)2)−3⋅49+5=3(x+23)2−43+153=3(x+23)2+113■\begin{align*} p(x) &= 3x^2 + 4x + 5 \\ &= 3\left(x^2 + \frac43x\right) + 5 \\ &= 3\left(x^2 + 2\cdot\frac23x + \left(\frac23\right)^2 - \left(\frac23\right)^2\right) + 5 \\ &= 3\left(x^2 + 2\cdot\frac23x + \left(\frac23\right)^2\right) - 3\cdot\frac{4}{9} + 5 \\ &= 3\left(x + \frac23\right)^2 - \frac{4}{3} + \frac{15}{3} \\ &= 3\left(x + \frac23\right)^2 + \frac{11}{3} \qed \end{align*}

The vertex is (−23,113)\left(-\frac23, \frac{11}{3}\right).

The second method is evaluating the quadratic at x=−b2ax = -\frac{b}{2a}.

Example: Convert 3x2+4x+53x^2 + 4x + 5 to vertex form.

a=3; b=4; c=5a = 3;\,b = 4;\, c = 5

Evaluate the function at x=−46=−23x = -\frac46 = -\frac23 results in

f(−23)=3(−23)2+4(−23)+5=43−83+153=113\begin{align*} f\left(-\frac23\right) &= 3\left(-\frac23\right)^2 + 4\left(-\frac23\right) + 5 \\ &= \frac{4}{3} - \frac{8}{3} + \frac{15}{3} \\ &= \frac{11}{3} \end{align*}

The answer is

3x2+4x+5=3(x+23)2+113■3x^2 + 4x + 5 = 3\left(x + \frac23\right)^2 + \frac{11}{3} \qed

The fact that the xx-value of the vertex is negative b2a\frac{b}{2a} comes from both the quadratic formula and also from setting the first derivative to zero.

 d dx(ax2+bx+c)=2ax+b=0\frac{\d}{\d x} \left(ax^2 + bx + c\right) = 2ax + b = 0

Discriminant

Δ=b2−4ac\Delta = b^2 - 4ac

The discriminant indicates the types of zeros. We assume real coefficients and a≠0a \neq 0.

case types of Roots
Δ>0{\Delta > 0} two distinct (different) real roots
Δ=0{\Delta = 0} two repeated (identical) real roots
Δ<0{\Delta < 0} SL: no real roots / HL: two complex conjugate roots

Properties

The vertex is the global maximum when a<0a < 0, and is the global minimum when a>0a > 0.

The function is symmetric about the vertex, meaning for any distance dd away from the vertex in the xx-direction

f(h+d)=f(h−d)f(h + d) = f(h - d)

Quadratic equations

Target Strategy
ax2+bx+c=0ax^2 + bx + c = 0 quadratic formula or factoring
(px+q)2=r(px + q)^2 = r x=1p(−q±r)\displaystyle x = \frac{1}{p}(-q \pm \sqrt r)

Unless the equation is already in the bottom form, always rearrange the equation into the top (standard) form before solving via factoring or quadratic formula.

The following are, or can be solved like, quadratics.

(3x−5)2=8−10x(3x - 5)^2 = 8 - 10x
4+x=2x−1\sqrt{4 + x} = 2x - 1
−26x+1+32x−5=3-\frac{2}{6x+1} + \frac{3}{2x - 5} = 3
3x2−x−35+2x=7\frac{3x^2 - x - 3}{5+2x} = 7

Quadratic inequalities

  1. Solve the quadratic f(x)=0f(x) = 0, where ff is some quadratic function.
  2. Determine if the graph looks like ∪\cup for a>0a > 0, or ∩\cap for a<0a < 0.
  3. With the visualization in mind (or on paper), determine the solutions.

Note: (px+q)2≥r(px + q)^2 \geq r is equivalent to y1=(px+q)2y_1 = (px + q)^2 and y2=ry_2 = r and y1≥y2y_1 \geq y_2.

Given that x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x + 2)(x + 3), we have

Inequality Solution
x2+5x+6>0x^2 + 5x + 6 > 0 x<−3,x>−2x < -3, x > -2
x2+5x+6<0x^2 + 5x + 6 < 0 −3<x<−2-3 < x < -2
−x2−5x−6>0-x^2 - 5x - 6 > 0 −3<x<−2-3 < x < -2
−x2−5x−6<0-x^2 - 5x - 6 < 0 x<−3,x>−2x < -3, x > -2

For ≤\leq and ≥\geq, change the inequality signs accordingly.

HL: Linear absolute value inequalities

See also absolute value functions

As surprising as it may sound, linear inequalities can involve quadratics.

Example: Solve ∣−2x−5∣>4\lvert -2x - 5\rvert > 4.

∣−2x−5∣>4(−2x−5)2>16(2x+5)2>16\begin{align*}\lvert -2x - 5\rvert &> 4 \\ (-2x - 5)^2 &> 16 \\ (2x + 5)^2 &> 16 \end{align*}

The equality is solved by

(2x+5)2=162x+5=±42x=−9,−1x=−92,−12\begin{align*} (2x + 5)^2 = 16 \\ 2x + 5 = \pm4 \\ 2x = -9, -1 \\ x = -\frac{9}{2}, -\frac{1}{2} \\ \end{align*}

Because graphs of absolute values of linear functions look like “V”, then the values of xx such that ∣−2x−5∣>4\lvert -2x - 5\rvert > 4 must be values less than some endpoint and values greater than some other endpoint. Therefore, the solution is

∈]−∞,−92[∪]−12,∞[\in \left]-\infty,-\frac{9}{2}\right[\cup \left]-\frac{1}{2},\infty\right[

or

x<−92,    x>−12■x < -\frac{9}{2},\;\; x > -\frac{1}{2}\qed

It is unnecessary, on the exam, to spend so many lines just to go from ∣−2x−5∣>4{\lvert -2x - 5\rvert > 4} to finding zeros of 2x+5=±4{2x + 5 = \pm4}. But it is conceptually useful to treat absolute value inequalities as quadratic inequalities.

If it is instead finding ∣f(x)∣<k\lvert f(x) \rvert < k then instead you need the single interval between the two roots of the equality.

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