Chapter A.6, Problem 113AYU

To determine

To find: The real solutions using the quadratic formula.

Answer to Problem 113AYU

$x=\frac{5+\sqrt{29}}{2}$ and $x=\frac{5-\sqrt{29}}{2}$

Explanation of Solution

$x^2-5x-1=0$

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

With $a=1, b=-5$, and $c=-1$, the discriminant is $b^2-4ac={(-5)}^2-4\left(1\right)(-1)=29$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=1, b=-5$, $c=-1$ and $b^2-4ac=29$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm \sqrt{29}}{2\left(1\right)}=\frac{5\pm \sqrt{29}}{2}$

$x=\frac{5+\sqrt{29}}{2}$ and $x=\frac{5-\sqrt{29}}{2}$