Question

Use the quadratic formula to solve the equation.

Answer 1

ANSWER

$x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4}$

EXPLANATION

The given quadratic equation is

$4 {x}^{2} - 10x + 5 = 0$

We compare this to

$a {x}^{2} + bx + c = 0$

to get a=4, b=-10, and c=5.

The quadratic formula is given by

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute these values into the formula to get:

$x = \frac{ - - 10 \pm \sqrt{ {( - 10)}^{2} - 4(4)(5)} }{2(4)}$

This implies that

$x = \frac{ 10 \pm \sqrt{ 100 - 80} }{8}$

$x = \frac{ 10 \pm \sqrt{ 20} }{8}$

$x = \frac{ 10 \pm2 \sqrt{ 5} }{8}$

$x = \frac{ 5 \pm \sqrt{ 5} }{4}$

The solutions are:

$x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4}$

Answer 2

Answer:

$\large\boxed{x=\dfrac{5-\sqrt5}{4},\ x=\dfrac{5+\sqrt5}{4}}$

Step-by-step explanation:

$\text{The quadratic formula for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac$

$\text{We have the equation:}\ 4x^2-10x+5=0\\\\a=4,\ b=-10,\ c=5\\\\b^2-4ac=(-10)^2-4(4)(5)=100-80=20>0\\\\x=\dfrac{-(-10)\pm\sqrt{20}}{2(4)}=\dfrac{10\pm\sqrt{4\cdot5}}{8}=\dfrac{10\pm\sqrt4\cdot\sqrt5}{8}=\dfrac{10\pm2\sqrt5}{8}\\\\=\dfrac{2(5\pm\sqrt5)}{8}=\dfrac{5\pm\sqrt5}{4}$