Chapter A.6, Problem 63AYU

To determine

To solve: $\left|x^2+x-1\right|=1$

Answer to Problem 63AYU

$x= -2, -1, 0, 1$

Explanation of Solution

$\left|x^2+x-1\right|=1$

$x^2+x-1=1$ or $x^2+x-1=-1$

$x^2+x-2=0$ or $x^2+x=0$

For $x^2+x-2=0$:

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=1, b= 1$, and $c=-2$, the discriminant is $b^2-4ac={\left(1\right)}^2-4\left(1\right)\left(-2\right)=9$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=1, b= 1, c=-2$ and $b^2-4ac=9$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-1\pm \sqrt{9}}{2}=\frac{-1\pm 3}{2}$

$x=\frac{-1+3}{2}=1$ or $x=\frac{-1-3}{2}=-2$

Therefore $x=1, -2$.

For $x^2+x=0$:

$x\left(x+1\right)=0$

$x=0, x+1=0$

$x=0, x=-1$

Hence the solutions are $0, -1, -2, 1$.