# Solve the equation?

Solve the equation

2x^2 + 8x - 1 = 0

by completing the square

### 7 Answers

- ?Lv 67 months ago
2x^2+8x-1=0, ie., 2x^2+8x-1+9=9, ie., 2(x^2+4x+4) =9, ie., (x+2)^2 =[(3/2)rt2]^2.

Then x = -2(+/-)(3/2)rt2, where rt2 = sqrt2.

- sepiaLv 77 months ago
2 x^2 + 8 x - 1 = 0

2/9 (x + 2)^2 = 1

Solutions:

x = -2 - 3/sqrt(2)

x = 3/sqrt(2) - 2

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- llafferLv 77 months ago
2x² + 8x - 1 = 0

Here's how I solve this by completing the square. Start with making the left half of the equation be in the form of (x² + bx). We'll add 1 to both sides then divide both sides by 2:

2x² + 8x = 1

x² + 4x = 1/2

Now that we have the form we need, the "complete the square" step is done by following these steps:

Start with x's coefficient (4)

half it (2)

square it (4)

Add that to both sides of the equation:

x² + 4x + 4 = 1/2 + 4

The left side is now a perfect square trinomial that can be factored and the right side can be simplified:

(x + 2)² = 1/2 + 8/2

(x + 2)² = 9/2

Square root of both sides:

x + 2 = ± √(9/2)

Subtract 2 from both sides, then simplify the radical, including rationalizing the denominator:

x = -2 ± √(9/2)

x = -2 ± 3/√2

x = -2 ± (3/2)√2

- AmyLv 77 months ago
Divide by the x^2 coefficient.

x^2 + 4x - 1/2 = 0

Complete the square

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

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

Solve for x

(x+2)^2 = 4.5

x+2 = ±√4.5

x+2 = ± 3√2/2

x = -2 ± 3√2/2

- ?Lv 77 months ago
2x² + 8x - 1 = 2(x² + 4x) - 1

i.e. 2(x + 2)² - 8 - 1

Hence, 2(x + 2)² - 9 = 0

so, 2(x + 2)² = 9

=> (x + 2)² = 9/2

so, x + 2 = ±3/√2 => ±3√2/2

i.e. x = -2 ± 3√2/2 => (-4 ± 3√2)/2

Hence, x = (-4 + 3√2)/2 or -(4 + 3√2)/2

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