Question

A quadratic equation is shown below:

Answer 1

PART  A. To determine how many solution the quadratic equation has by just  determining the radicand, we are going to use the discriminant of the quadratic formula. Remember that the discriminant of the quadratic formula is $b^2-4ac$. If the discriminant is $\ \textgreater \ 0$, the quadratic equation will have tow real solutions. If the discriminant is $\ \textless \ 0$, the quadratic equation will have no real solution. If the discriminant is equal to zero, the quadratic equation will have one number real solution (double). We now for our equation that $a=16$, $b=-8$, and $c=1$. Lest replace those values in our discriminant:
$b^2-4ac$
$(-8)^2-4(16)(1)$
$64-64=0$

Since the discriminant of our quadratic equation is zero, we can conclude that it will have one number real solution (with duplicity).

PART B. Here we are going to use the quadratic formula because even though the process will be longer, it will easier than finding common factors. Remember that the quadratic formula is: $x= \frac{-b(+ or -) \sqrt{b^2-4ac} }{2a}$. From our quadratic equation we can infer that $a=4$, $b=-8$, and $c=45$. Lets replace those values in our formula:
$x= \frac{-(-8)(+ or -) \sqrt{(-8)^2-4(4)(-45)} }{2(8)}$
$x= \frac{8(+ or -) \sqrt{64+720)} }{8}$
$x= \frac{8(+ or -) \sqrt{784} }{8}$
$x= \frac{8(+or-)28}{8}$
$x= \frac{8+28}{8}$ or $x= \frac{8-28}{8}$
$x= \frac{9}{2}$ or $x=- \frac{5}{2}$

We can conclude that the solutions of our quadratic equation are tex]x= \frac{9}{2} [/tex] and $x=- \frac{5}{2}$.