Quadratic functions and their roots

The qua­dratic func­tion is the poly­no­mial of degree $2$, and it is the most impor­tant func­tion of this chap­ter. Its graph is a curve called a parabola, and the shape and posi­tion of that curve decide how to solve qua­dratic inequa­tions later on. Before draw­ing it, you need a firm hold on its roots: how to find them, how to tell what kind they are with­out find­ing them, and how their sum, prod­uct and dif­fer­ence con­nect to the coef­fi­cients.

Qua­dratic func­tion and qua­dratic equa­tion#

A func­tion of the form

$$y = f(x) = ax^2 + bx + c \quad (a \neq 0)$$

is called a qua­dratic func­tion, or a qua­dratic expres­sion.

The equa­tion

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

is called a qua­dratic equa­tion. A value of $x$ that sat­is­fies it is called a root of the equa­tion. A root of the equa­tion is also called a zero of the func­tion, because the func­tion takes the value $0$ there.

The con­di­tion $a \neq 0$ mat­ters: if $a = 0$ the $x^2$ term dis­ap­pears and we are back to a lin­ear func­tion.

The roots and the dis­crim­i­nant#

The roots of the qua­dratic equa­tion $ax^2 + bx + c = 0$ are

$$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}$$

where

$$\Delta = b^2 - 4ac$$

is called the dis­crim­i­nant. The dis­crim­i­nant decides the kind of roots, because it sits under the square root sign.

• $\Delta > 0$: the roots are real and dis­tinct (two dif­fer­ent real roots).
• $\Delta = 0$: the roots are real and equal (both are $\displaystyle -\frac{b}{2a}$).
• $\Delta < 0$: there are no real roots. The roots are imag­i­nary, because the square root of a neg­a­tive num­ber is not a real num­ber.

Exam­ple 1#

Find the nature of the roots of (a) $2x^2 - 7x + 3 = 0$ (b) $x^2 - 6x + 9 = 0$ (c) $x^2 + x + 1 = 0$. Solve those that have real roots.

(a) Here $a = 2$, $b = -7$, $c = 3$. So $\Delta = 49 - 24 = 25 > 0$: real and dis­tinct roots.

$$\displaystyle x = \frac{7 \pm 5}{4}, \quad \text{so } x = 3 \text{ or } x = \frac{1}{2}$$

(b) Here $a = 1$, $b = -6$, $c = 9$. So $\Delta = 36 - 36 = 0$: real and equal roots, both equal to $\displaystyle \frac{6}{2} = 3$.

(c) Here $a = b = c = 1$. So $\Delta = 1 - 4 = -3 < 0$: no real roots.

Sum, prod­uct and dif­fer­ence of the roots#

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, then

1. Sum of roots: $\displaystyle \alpha + \beta = -\frac{b}{a} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
2. Prod­uct of roots: $\displaystyle \alpha\beta = \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
3. Dif­fer­ence of roots: $\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{b^2 - 4ac}}{|a|}$

The third result is very impor­tant and is easy to prove from the first two. Start from the iden­tity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$:

$$\displaystyle \begin{aligned} (\alpha - \beta)^2 &= \left(-\frac{b}{a}\right)^2 - \frac{4c}{a} \\ &= \frac{b^2}{a^2} - \frac{4ac}{a^2} \\ &= \frac{b^2 - 4ac}{a^2} \end{aligned}$$

Tak­ing the pos­i­tive square root of both sides,

$$\displaystyle |\alpha - \beta| = \frac{\sqrt{b^2 - 4ac}}{|a|}$$

The mod­u­lus signs appear because a square root is never neg­a­tive, and we do not know in advance which root is larger or whether $a$ is pos­i­tive.

Exam­ple 2#

With­out solv­ing, find the sum, prod­uct and dif­fer­ence of the roots of $3x^2 + 5x - 2 = 0$. Then check by solv­ing.

Here $a = 3$, $b = 5$, $c = -2$.

$$\displaystyle \alpha + \beta = -\frac{5}{3}, \qquad \alpha\beta = \frac{-2}{3} = -\frac{2}{3}$$

$$\displaystyle \Delta = 25 - 4(3)(-2) = 25 + 24 = 49, \qquad |\alpha - \beta| = \frac{\sqrt{49}}{3} = \frac{7}{3}$$

Check: $\displaystyle x = \frac{-5 \pm 7}{6}$ gives $\displaystyle x = \frac{1}{3}$ or $x = -2$. Their sum is $\displaystyle \frac{1}{3} - 2 = -\frac{5}{3}$, their prod­uct is $\displaystyle -\frac{2}{3}$, and their dif­fer­ence is $\displaystyle \frac{1}{3} + 2 = \frac{7}{3}$. All three agree.

Exam­ple 3#

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

Here $\alpha + \beta = 5$ and $\alpha\beta = 3$.

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19$$

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 25 - 12 = 13$$

Nei­ther answer needed the roots them­selves, which are the untidy num­bers $\displaystyle \frac{5 \pm \sqrt{13}}{2}$.

The fac­tored form#

Result. If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, then

$$ax^2 + bx + c = a(x - \alpha)(x - \beta)$$

Proof. Take $a$ out as a com­mon fac­tor and use the sum and prod­uct of the roots:

$$\displaystyle \begin{aligned} ax^2 + bx + c &= a\left[x^2 + \frac{b}{a}x + \frac{c}{a}\right] \\ &= a\left[x^2 - \left(-\frac{b}{a}\right)x + \frac{c}{a}\right] \\ &= a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] \\ &= a\left[x^2 - \alpha x - \beta x + \alpha\beta\right] \\ &= a(x - \alpha)(x - \beta) \end{aligned}$$

This is called the fac­tored form of the qua­dratic. You will use it again and again when draw­ing graphs and solv­ing inequa­tions, because it shows the roots at a glance.

Exam­ple 4#

Write $2x^2 - 7x + 3$ in fac­tored form.

From Exam­ple 1(a), the roots are $3$ and $\displaystyle \frac{1}{2}$, and $a = 2$. So

$$\displaystyle 2x^2 - 7x + 3 = 2(x - 3)\left(x - \frac{1}{2}\right) = (x - 3)(2x - 1)$$

Mul­ti­ply­ing out $(x - 3)(2x - 1)$ gives $2x^2 - x - 6x + 3 = 2x^2 - 7x + 3$, as it should.

Exam­ple 5#

For what val­ues of $k$ does $x^2 + kx + 16 = 0$ have equal roots?

Equal roots need $\Delta = 0$: $k^2 - 4(1)(16) = 0$, so $k^2 = 64$ and $k = 8$ or $k = -8$.

Prac­tice#

1. Find the dis­crim­i­nant of $x^2 - 4x + 1 = 0$ and state the nature of its roots.
2. Find the nature of the roots of $4x^2 + 12x + 9 = 0$, and find the root.
3. Show that $2x^2 - 3x + 5 = 0$ has no real roots.
4. Solve $6x^2 - x - 2 = 0$ using the for­mula.
5. Write the sum and the prod­uct of the roots of $5x^2 - 3x - 7 = 0$.
6. Find the dif­fer­ence of the roots of $x^2 - 7x + 10 = 0$ with­out solv­ing it.
7. Find the dif­fer­ence of the roots of $2x^2 + 3x - 4 = 0$.
8. Write $3x^2 - 10x + 3$ in fac­tored form.
9. If $\alpha$ and $\beta$ are the roots of $x^2 - 4x + 2 = 0$, find $\alpha^2 + \beta^2$ and $|\alpha - \beta|$.
10. For what val­ues of $k$ does $x^2 - kx + 9 = 0$ have equal roots?

Answers#