# How do you solve 0= 4x ^ { 2} + 8x + 1?

Mar 20, 2018

$x = - 1 \pm \frac{1}{2} \sqrt{3}$

#### Explanation:

$\text{given a quadratic equation in "color(blue)"standard form}$

•color(white)(x)ax^2+bx+c=0color(white)(x);a!=0

$\text{we can solve for x using the "color(blue)"quadratic formula}$

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

$0 = 4 {x}^{2} + 8 x + 1 \text{ is in standard form}$

$\text{with "a=4,b=8" and } c = 1$

$\Rightarrow x = \frac{- 8 \pm \sqrt{64 - 16}}{8}$

$\textcolor{w h i t e}{\Rightarrow x} = \frac{- 8 \pm \sqrt{48}}{8} = \frac{- 8 \pm 4 \sqrt{3}}{8}$

$\Rightarrow x = - 1 \pm \frac{1}{2} \sqrt{3} \leftarrow \textcolor{red}{\text{exact solutions}}$

Mar 20, 2018

$x = - 1 + \frac{1}{2} \sqrt{3} , - 1 - \frac{1}{2} \sqrt{3}$

#### Explanation:

$a {x}^{2} + b x + c = 0$ (quadratic form)
$\left(4\right) {x}^{2} + \left(8\right) x + \left(1\right) = 0$ (your equation)

So:
$a = 4$
$b = 8$
$c = 1$

Now plug those in to the quadratic formula and solve:
$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

 x = (-(8) \pm sqrt(8^2-4(4)(1))) / (2(4)

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

$x = \frac{- 8 \setminus \pm \sqrt{48}}{8}$

Since $48$ isn't an even square root, we will factor:

$x = \frac{- 8 \setminus \pm \sqrt{48}}{8} \rightarrow x = \frac{- 8 \setminus \pm \sqrt{3 \cdot {4}^{2}}}{8} \rightarrow x = \frac{- 8 \setminus \pm 4 \sqrt{3}}{8}$

Now divide by $4$ to simplify:

$\frac{\cancel{\text{-8}} \textcolor{red}{2} \setminus \pm \cancel{4} \textcolor{red}{1} \sqrt{3}}{\cancel{8} \textcolor{red}{2}} \rightarrow \frac{- 2 \setminus \pm 1 \sqrt{3}}{2} \rightarrow - 1 \setminus \pm \frac{1}{2} \sqrt{3}$

$x = - 1 + \frac{1}{2} \sqrt{3} , - 1 - \frac{1}{2} \sqrt{3}$

Mar 20, 2018

(-2±√3)/2

#### Explanation:

We will solve this quadratic equation using quadratic formula, which goes as follows :-
4$x$²+8$x$+1=0
In this equation
a=4
b=8
c=1
First we will find the discriminant(D) for this equation
D=b²-4ac
D= 8²-4×4×1
D=64-16
D=48
The solution of this equation according to quadratic formula is given as
(-b± √D) /2a
Now let the two solutions of this equation be A and B
Therefore,
A=(-b-√D) /2a and B=(-b+√D) /2a
A=(-8-√48)/2×4 and B=(-8+√48) /2×4
A=(-8-√48)/8 and B=(-8+√48)/8
A=(-8-4√3)/8 and B=(-8+4√3)/8
A=(-2-√3)/2 and B=(-2+√3)/2