Chapter A.6, Problem 115AYU

To determine

To find: The real solutions using the quadratic formula: $2x^2-5x+3=0$.

Answer to Problem 115AYU

$x=\frac{3}{2},-1$

Explanation of Solution

$2x^2-5x+3=0$

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

With $a=2, b=-5$, and $c=3$, the discriminant is $b^2-4ac={(-5)}^2-4\left(2\right)\left(3\right)=1$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=2, b=-5$, and $c=3$ and $b^2-4ac=1$.

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

$x=\frac{5-1}{4}=1$ and $x=\frac{5+1}{4}=\frac{6}{4}=\frac{3}{2}$

Therefore $x=\frac{3}{2},1$.