Chapter A.6, Problem 111AYU

To determine

To find: The real solutions. Use the quadratic formula: $x^2-4x+2=0$.

Answer to Problem 111AYU

$x=2+\sqrt{2}, 2-\sqrt{2}$

Explanation of Solution

$x^2-4x+2=0$

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

With $a=1, b=-4$, and $c=2$, the discriminant is $b^2-4ac={(-4)}^2-4\left(1\right)\left(2\right)=8$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=1, b=-4, c=2$ and $b^2-4ac=8$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{8}}{2\left(1\right)}=\frac{4\pm 2\sqrt{2}}{2}$

$x=\frac{4+2\sqrt{2}}{2}=2+\sqrt{2}$ and $x=\frac{4-2\sqrt{2}}{2}=2-\sqrt{2}$

Therefore $x=2+\sqrt{2}, 2-\sqrt{2}$.