Question

$x^2 + bx + c = 0$
in the equation above, $b$ and $c$ are constants, $-b + \sqrt{b^2 - 4c} = 8$, and $-b - \sqrt{b^2 - 4c} = 2$. if $x$ is a solution of the equation, what is one possible value of $x$?

Explanation:

Step1: Recall quadratic formula

The quadratic formula for \(ax^2 + bx + c = 0\) (here \(a = 1\)) is \(x=\frac{-b\pm\sqrt{b^2 - 4c}}{2}\). We are given \(-b+\sqrt{b^2 - 4c}=8\) and \(-b-\sqrt{b^2 - 4c}=2\).

Step2: Find the two solutions

Let \(x_1=\frac{-b + \sqrt{b^2 - 4c}}{2}\) and \(x_2=\frac{-b-\sqrt{b^2 - 4c}}{2}\). Substitute the given values:
For \(x_1\): \(x_1=\frac{8}{2}=4\)
For \(x_2\): \(x_2=\frac{2}{2}=1\)

Answer:

1 or 4 (either 1 or 4 is a correct answer, here we can write 1 or 4, for example, 1)