# How do you solve the quadratic equation 4x^2 - 8x + 1= 0 by using the Quadratic Formula?

Apr 10, 2016

Using quadratic formula: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

#### Explanation:

you assume the value
=> 4 to be your ''a''
=> -8 to be your ''b''
=> 1 to be your ''c''

Now just replace the values in the formula, and obtain your two values

-> N.B You will obtain two values (might be one positive one and the other, a negative one)

Apr 10, 2016

$x = \frac{2 \pm \sqrt{3}}{2}$

#### Explanation:

color(blue)(4x^2-8x+1=0

This is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Note: $a \mathmr{and} b$ are the coefficients of ${x}^{2} \mathmr{and} x$ and $c$ is the constant number

So,

color(purple)(a=4,b=-8,c=1

$\rightarrow x = \frac{- \left(- 8\right) \pm \sqrt{- {8}^{2} - 4 \left(4\right) \left(1\right)}}{2 \left(4\right)}$

$\rightarrow x = \frac{8 \pm \sqrt{64 - \left(16\right)}}{8}$

$\rightarrow x = \frac{8 \pm \sqrt{48}}{8}$

$\rightarrow x = \frac{8 \pm \sqrt{16 \cdot 3}}{8}$

$\rightarrow x = \frac{8 \pm 4 \sqrt{3}}{8}$

$\rightarrow x = \frac{{\cancel{8}}^{2} \pm {\cancel{4}}^{1} \sqrt{3}}{\cancel{8}} ^ 2$

color(green)(rArrx=(2+-sqrt3)/2

Remember that $\pm$ means "plus or minus"

Which implies that

color(indigo)(x=((2+sqrt3)/(2),(2-sqrt3)/(2))

If you are a bit confused with the Quadratic formula

Watch this video