# How do you solve 4x^2+8x+3=0 by completing the square?

##### 2 Answers
May 24, 2017

$x = - \frac{1}{2} \mathmr{and} x = - \frac{3}{2}$

#### Explanation:

$4 {x}^{2} + 8 x + 3 = 0$
$4 {x}^{2} + 8 x + 4 = - 3 + 4$
$4 \left({x}^{2} + 2 x + 1\right) = 1$
${\left(x + 1\right)}^{2} = \frac{1}{4}$
$x + 1 = \pm \frac{1}{2}$
$x = - 1 \pm \frac{1}{2}$
$x 1 = - \frac{1}{2}$
$x 2 = - \frac{3}{2}$

May 24, 2017

Answer: $x = - \frac{1}{2}$ and $x = - \frac{3}{2}$

#### Explanation:

Step 1. Factor out any numbers in front of the ${x}^{2}$ term

$4 \left({x}^{2} + 2 x + \frac{3}{4}\right) = 0$

Step 2. Cut the term in front of $x$ in half, then square it. Add and subtract the result inside the parenthesis

• Term in front of $x$: 2
• Cut it in half: $1$
• Square it: ${1}^{2} = 1$
• Add and subtract that inside the parenthesis

$4 \left({x}^{2} + 2 x + 1 - 1 + \frac{3}{4}\right) = 0$

Step 3. Isolate and rewrite your perfect square

• Isolate the perfect square
$4 \left(\left({x}^{2} + 2 x + 1\right) - 1 + \frac{3}{4}\right) = 0$

• Rewrite the perfect square
$4 \left({\left(x + 1\right)}^{2} - 1 + \frac{3}{4}\right) = 0$

• Simplify the remaining addition / subtraction term
$4 \left({\left(x + 1\right)}^{2} - \frac{1}{4}\right) = 0$

Step 4. Multiply the coefficient term back through
$4 {\left(x + 1\right)}^{2} - 1 = 0$

Step 5. Solve the equation

• Add $1$ to both sides
$4 {\left(x + 1\right)}^{2} = 1$

• Divide both sides by $4$
${\left(x + 1\right)}^{2} = \frac{1}{4}$

• Square root both sides
$x + 1 = \pm \sqrt{\frac{1}{4}}$

• Subtract $1$ from both sides
$x = - 1 \pm \frac{1}{2}$

Answer: $x = - \frac{1}{2}$ and $x = - \frac{3}{2}$