Chapter A.6, Problem 119AYU

To determine

To find: The real solutions using the quadratic formula: $4x^2=1-2x$.

Answer to Problem 119AYU

$x=\frac{-1+\sqrt{5}}{4}$ and $x=\frac{-1-\sqrt{5}}{4}$

Explanation of Solution

$4x^2=1-2x$

$4x^2+2x-1=0$

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

With $a=4, b=2$, $c=-1$, the discriminant is $b^2-4ac={\left(2\right)}^2-4\left(4\right)\left(-1\right)=20$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=4, b=2$, $c=-1$ and $b^2-4ac=20$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(2\right)\pm \sqrt{20}}{2\left(4\right)}=\frac{-2\pm 2\sqrt{5}}{8}=\frac{-1\pm \sqrt{5}}{4}$

$x=\frac{-1+\sqrt{5}}{4}$ and $x=\frac{-1-\sqrt{5}}{4}$