How do you solve 8x+1=4x^2 by quadratic formula?

Feb 7, 2016

$8 x + 1 = 4 {x}^{2}$

$\rightarrow 8 x = 4 {x}^{2} - 1$

$\rightarrow 4 {x}^{2} - 1 - 8 x = 0$

In standard form:

$\rightarrow 4 {x}^{2} - 8 x - 1 = 0$

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

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

In this case $a = 4 , b = - 8 , c = - 1$

Substitute the values into the formula:

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

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

$\rightarrow x = \frac{8 \pm \sqrt{64 + 16}}{8}$

$\rightarrow x = \frac{8 \pm \sqrt{80}}{8}$

$\rightarrow x = \frac{8 \pm \sqrt{16 \cdot 5}}{8}$

$\rightarrow x = \frac{8 \pm 4 \sqrt{5}}{8}$

$\rightarrow x = \frac{2 \pm \sqrt{5}}{2}$

Feb 7, 2016

$\frac{2 \pm \sqrt{5}}{2}$

Explanation:

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

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

so $8 x + 1 = 4 {x}^{2}$

$\rightarrow 4 {x}^{2} - 8 x - 1 = 0$

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

$\rightarrow \frac{8 \pm \sqrt{64 + 16}}{8}$

$\rightarrow \frac{8 \pm \sqrt{80}}{8}$

$\rightarrow \frac{8 \pm 4 \sqrt{5}}{8}$

$\rightarrow \frac{8 \pm 4 \sqrt{5}}{8}$

$\rightarrow \frac{2 \pm \sqrt{5}}{2}$