Question

Giving 40 points!!! please help

Answer 1

Answer:

$\textsf{D.} \quad x=4 \pm \sqrt{17}$

Step-by-step explanation:

Rearrange the equation so it equals zero:

$\begin{aligned} x^2+4 & =8x+5 \\ \implies x^2+4-8x-5 & = 0\\x^2-8x-1&=0\end{aligned}$

Now use the quadratic formula to solve for x:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$a=1, \quad b=-8, \quad c=-1$

$\implies x=\dfrac{-(-8) \pm \sqrt{(-8)^2-4(1)(-1)}}{2(1)}$

$\implies x=\dfrac{8 \pm \sqrt{68}}{2}$

$\implies x=\dfrac{8 \pm \sqrt{4\cdot 17}}{2}$

$\implies x=\dfrac{8 \pm \sqrt{4}\sqrt{17}}{2}$

$\implies x=\dfrac{8 \pm 2\sqrt{17}}{2}$

$\implies x=4 \pm \sqrt{17}$