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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