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Solve x^2 - 8x + 52 by Completing the Square

Learn how to solve x^2 - 8x + 52 = 0 by completing the square, forming (x - 4)^2 = -36 and finding the complex roots 4 ± 6i.

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Problem

Solve by completing the square:

x28x+52=0x^2 - 8x + 52 = 0

Step 1: Isolate the xx Terms

To complete the square, move the constant term away from the x2x^2 and xx terms. Subtract 5252 from both sides:

x28x=52x^2 - 8x = -52

Step 2: Create a Perfect Square

Take half of the coefficient of xx and square it. Half of 8-8 is 4-4, and

(4)2=16(-4)^2 = 16

Add 1616 to both sides:

x28x+16=52+16x^2 - 8x + 16 = -52 + 16

x28x+16=36x^2 - 8x + 16 = -36

Step 3: Rewrite the Trinomial

The left side is now a perfect square trinomial:

x28x+16=(x4)2x^2 - 8x + 16 = (x - 4)^2

So the equation becomes

(x4)2=36(x - 4)^2 = -36

Step 4: Take Square Roots

Take the square root of both sides, remembering both the positive and negative possibilities:

x4=±36x - 4 = \pm \sqrt{-36}

Since

36=6i\sqrt{-36} = 6i

we get

x4=±6ix - 4 = \pm 6i

Step 5: Solve for xx

Add 44 to both sides:

x=4±6ix = 4 \pm 6i

So the solutions are

x=46iandx=4+6ix = 4 - 6i \quad \text{and} \quad x = 4 + 6i

Concepts

Quadratic Formula and Completing the Square

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