Question

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

Answer 1

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

Explanation:

Hello,

step 1 - we divide all terms by 2

`2x^2+8x-1=0 <=> x^2+4x-(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-(1)/(2)=0<=>(x+2)^2-4-(1)/(2)=0\\`

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

`(x+2)^2-4-(1)/(2)=0\\\\<=> (x+2)^2=4+(1)/(2)=(8+1)/(2)=(9)/(2)`

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

`x+2=\sqrt{(9)/(2)}`

or

`x+2=-\sqrt{(9)/(2)}`

step 5 - we subtract 2 from both sides

`x+2=\sqrt{(9)/(2)}<=> x=(3)/(√(2))-2=0.12132...`

or

`x+2=-\sqrt{(9)/(2)}<=> x=-(3)/(√(2))-2=-4.12132...`

so the solutions are 0.12 and -4.12

hope this helps