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

Question

25075 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-24T00:29:41

$x = - 1/2 and x = - 3/2$

$4x^2 + 8x + 3 = 0$
$4x^2 + 8x + 4 = - 3 + 4$
$4(x^2 + 2x + 1) = 1$
$(x + 1)^2 = 1/4$
$x + 1 = +- 1/2$
$x = - 1 +- 1/2$
$x1 = - 1/2$
$x2 = - 3/2$

Answer 2 · 2017-05-24T00:42:02

Answer: $x=-1/2$ and $x=-3/2$

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

$4(x^2+2x+3/4)=0$

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

$4(x^2+2x+1-1+3/4)=0$

Step 3. Isolate and rewrite your perfect square

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

Step 5. Solve the equation

Answer: $x=-1/2$ and $x=-3/2$

How do you solve 4x^2+8x+3=04x^2+8x+3=0 by completing the square?