Question

Solve the equation 2x^2+8x-1=o by completing the square. Give your answer to 2 decimal places

Answer 1

Answer:

$\boxed{\sf \ \ \ x=-4.12 \ or \ x=0.12 \ \ \ }$

Step-by-step explanation:

Hello,

step 1 - we divide all terms by 2

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

step 2 - we complete the square

we can notice that

$x^2+4x=(x+2)^2-4$

so

$2x^2+8x-1=0 x^2+4x-\dfrac{1}{2}=0(x+2)^2-4-\dfrac{1}{2}=0$

step 3 - we move the constant term to the right of the equation

$(x+2)^2-4-\dfrac{1}{2}=0\\\\ (x+2)^2=4+\dfrac{1}{2}=\dfrac{8+1}{2}=\dfrac{9}{2}$

step 4 - we take the square root on both sides of the equation

$x+2=\sqrt{\dfrac{9}{2}}$

or

$x+2=-\sqrt{\dfrac{9}{2}}$

step 5 - we subtract 2 from both sides

$x+2=\sqrt{\dfrac{9}{2}} x=\dfrac{3}{\sqrt{2}}-2=0.12132...$

or

$x+2=-\sqrt{\dfrac{9}{2}} x=-\dfrac{3}{\sqrt{2}}-2=-4.12132...$

so the solutions are 0.12 and -4.12

hope this helps