Solve by Completing the Square 2x^2+7x-1=0

Problem

2*x2+7*x−1=0

Solution

1. Move the constant to the right side of the equation by adding 1 to both sides.

2*x2+7*x=1

2. Divide by the leading coefficient to ensure the x2 term has a coefficient of 1

x2+7/2*x=1/2

3. Calculate the value to complete the square by taking half of the x coefficient and squaring it: (1/2⋅7/2)2=(7/4)2=49/16

Value=49/16

4. Add the value to both sides of the equation to maintain equality.

x2+7/2*x+49/16=1/2+49/16

5. Simplify the right side by finding a common denominator: 1/2=8/16 so 8/16+49/16=57/16

x2+7/2*x+49/16=57/16

6. Factor the perfect square trinomial on the left side.

(x+7/4)2=57/16

7. Take the square root of both sides, remembering to include the ± sign.

x+7/4=±√(,57/16)

8. Simplify the radical on the right side.

x+7/4=±√(,57)/4

9. Isolate the variable by subtracting 7/4 from both sides.

x=−7/4±√(,57)/4

Final Answer

x=(−7±√(,57))/4

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