Question

which are the solutions of $x^2 = -7x - 8$?
$\frac{7}{2} - \frac{\sqrt{17}}{4}, \frac{7}{2} + \frac{\sqrt{17}}{4}$
$-\frac{7}{2} - \frac{\sqrt{17}}{4}, -\frac{7}{2} + \frac{\sqrt{17}}{4}$
$\frac{7 - \sqrt{17}}{2}, \frac{7 + \sqrt{17}}{2}$
$\frac{-7 - \sqrt{17}}{2}, \frac{-7 + \sqrt{17}}{2}$

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation \(x^{2}=-7x - 8\) in standard quadratic form \(ax^{2}+bx + c = 0\). Add \(7x\) and \(8\) to both sides: \(x^{2}+7x + 8=0\). Here, \(a = 1\), \(b = 7\), \(c = 8\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b = 7\), \(c = 8\) into the formula. First, calculate the discriminant \(\Delta=b^{2}-4ac=(7)^{2}-4\times1\times8 = 49 - 32=17\). Then, \(x=\frac{-7\pm\sqrt{17}}{2\times1}=\frac{-7\pm\sqrt{17}}{2}\). So the solutions are \(\frac{-7 - \sqrt{17}}{2}\) and \(\frac{-7+\sqrt{17}}{2}\).

Answer:

\(\boldsymbol{\frac{-7 - \sqrt{17}}{2},\frac{-7+\sqrt{17}}{2}}\) (corresponding to the last option: \(\frac{-7 - \sqrt{17}}{2},\frac{-7+\sqrt{17}}{2}\))