# How do you write a quadratic equation in standard form with solutions -2 and 7?

Sep 9, 2016

$\textcolor{g r e e n}{{x}^{2} - 5 x - 14 = 0}$

#### Explanation:

If color(red)("(-2)) and color(blue)(""(7)) are solutions to a quadratic equation $\textcolor{g r e e n}{f \left(x\right) = 0}$

then (x-color(red)(""(-2))) and $\left(x - \textcolor{b l u e}{7}\right)$ are factors of color(green)(f(x)
and we can write
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{f \left(x\right)} = \left(x + 2\right) \left(x - 7\right) = {x}^{2} - 5 x - 14$

Sep 9, 2016

$a \left({x}^{2} - 5 x - 14\right) = 0 , a \ne 0$

#### Explanation:

Let the reqd. quadr. eqn. be $p \left(x\right) = a {x}^{2} + b x + c = 0 , a \ne 0$.

$x = - 2 \text{ is a soln. } \Rightarrow p \left(- 2\right) = 0 \Rightarrow 4 a - 2 b + c = 0. . . \ldots \left(1\right)$

Similarly, $p \left(7\right) = 0 \Rightarrow 49 a + 7 b + c = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

$\left(2\right) - \left(1\right) \Rightarrow 45 a + 9 b = 0 \Rightarrow b = - 5 a \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$

Sub.ing this $b$ in $\left(1\right) , 4 a + 10 a + c = 0 \Rightarrow c = - 14 a \ldots \ldots \ldots \left(4\right)$.

Using $\left(3\right) \mathmr{and} \left(4\right)$, we get the reqd. quadr.

$p \left(x\right) = a {x}^{2} - 5 a x - 14 a = 0 , \mathmr{and} , p \left(x\right) = a \left({x}^{2} - 5 x - 14\right) = 0 , a \ne 0$