Question

for what value of n does the quadratic equation x² + 2x + n = 0 have solutions of x = 4 and x = -2?
a -8
b -2
c 2
d 6
e 8

Explanation:

Step1: Substitute x = 4 into equation

$4^{2}+2\times4 + n=0$
$16 + 8 + n=0$
$24 + n=0$

Step2: Solve for n

$n=-24$ (This is wrong, let's use the correct relationship). If $x_1$ and $x_2$ are the roots of the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b = 2$, $c = n$), then by Vieta's formulas, $x_1\times x_2=\frac{c}{a}$. Given $x_1 = 4$ and $x_2=-2$, and $a = 1$, $c = n$.

Step3: Apply Vieta's formula

$4\times(-2)=\frac{n}{1}$
$n=-8$

Answer:

A. - 8