# 10.8 Construct a Quadratic Equation from its Roots

It is possible to construct an equation from its roots, and the process is surprisingly simple. Consider the following:

Example 10.8.1

Construct a quadratic equation whose roots are $x = 4$ and $x = 6$.

This means that $x = 4$ (or $x - 4 = 0$) and $x = 6$ (or $x - 6 = 0$).

The quadratic equation these roots come from would have as its factored form:

$(x - 4)(x - 6) = 0$

All that needs to be done is to multiply these two terms together:

$(x - 4)(x - 6) = x^2 - 10x + 24 = 0$

This means that the original equation will be equivalent to $x^2 - 10x + 24 = 0$.

This strategy works for even more complicated equations, such as:

Example 10.8.2

Construct a polynomial equation whose roots are $x = \pm 2$ and $x = 5$.

This means that $x = 2$ (or $x - 2 = 0$), $x = -2$ (or $x + 2 = 0$) and $x = 5$ (or $x - 5 = 0$).

These solutions come from the factored polynomial that looks like:

$(x - 2)(x + 2)(x - 5) = 0$

Multiplying these terms together yields:

$\begin{array}{rrrrcrrrr} &&(x^2&-&4)(x&-&5)&=&0 \\ x^3&-&5x^2&-&4x&+&20&=&0 \end{array}$

The original equation will be equivalent to $x^3 - 5x^2 - 4x + 20 = 0$.

Caveat: the exact form of the original equation cannot be recreated; only the equivalent. For example, $x^3 - 5x^2 - 4x + 20 = 0$ is the same as $2x^3 - 10x^2 - 8x + 40 = 0$, $3x^3 - 15x^2 - 12x + 60 = 0$, $4x^3 - 20x^2 - 16x + 80 = 0$, $5x^3 - 25x^2 - 20x + 100 = 0$, and so on. There simply is not enough information given to recreate the exact original—only an equation that is equivalent.

# Questions

Construct a quadratic equation from its solution(s).

1. 2, 5
2. 3, 6
3. 20, 2
4. 13, 1
5. 4, 4
6. 0, 9
7. $\dfrac{3}{4}, \dfrac{1}{4}$
8. $\dfrac{5}{8}, \dfrac{5}{7}$
9. $\dfrac{1}{2}, \dfrac{1}{3}$
10. $\dfrac{1}{2}, \dfrac{2}{3}$
11. ± 5
12. ± 1
13. $\pm \dfrac{1}{5}$
14. $\pm \sqrt{7}$
15. $\pm \sqrt{11}$
16. $\pm 2\sqrt{3}$
17. 3, 5, 8
18. −4, 0, 4
19. −9, −6, −2
20. ± 1, 5
21. ± 2, ± 5
22. $\pm 2\sqrt{3}, \pm \sqrt{5}$