Question

Suppose the quadratic function $f\left(x\right)=a{x}^2+bx+c$ is such that $f(-2)=0and\frac{-b}{2a}=1$. Solve $f\left(x\right)=0$

• A. $x=-2,4$
• B. $x=2,-4$
• C. $x=2,6$
• D. $x=-2,-4$

Answer 1

The correct option is A $x=-2,4$

Given,

$f\left(x\right)=a{x}^2+bx+c$

$f(-2)=0$

$\Rightarrow f(-2)=a(-2{)}^2+b(-2)+c=0$

$\Rightarrow 4a-2b+c=0$

but, $-\frac{b}{2a}=1\Rightarrow b=-2a$

$\Rightarrow 4a-2(-2a)+c=0$

$\therefore c=-8a$

Now,

$f\left(x\right)=0$

$a{x}^2+bx+c=0$

$\Rightarrow x_1,x_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$=1\pm \frac{\sqrt{(-2a{)}^2-4a(8a)}}{2a}$

$=1\pm \frac{6a}{2a}$

$=1\pm 3$

$\therefore x_1,x_2=-2,4$