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Solving x2+10x+4=15x^2 + 10x + 4 = 15 by Completing the Square

Solve x2+10x+4=15x^2 + 10x + 4 = 15 by completing the square, demonstrating the procedure when constants appear on both sides of the equation.

Step 1 — Isolate the variable terms. The variable terms x2x^2 and 10x10x are already on the left, but the constant 44 must be moved to the right. Subtract 44 from both sides:

x2+10x=11x^2 + 10x = 11

Step 2 — Find (12b)2\left(\frac{1}{2} \cdot b\right)^2 and add it to both sides. The coefficient of xx is 1010, so b=10b = 10. Compute: (12(10))2=52=25\left(\frac{1}{2}(10)\right)^2 = 5^2 = 25. Add 2525 to both sides:

x2+10x+25=11+25x^2 + 10x + 25 = 11 + 25

Step 3 — Factor the perfect square trinomial. The left side factors as a binomial square:

(x+5)2=36(x + 5)^2 = 36

Step 4 — Apply the Square Root Property:

x+5=±36x + 5 = \pm\sqrt{36}

Step 5 — Simplify the radical and solve. Since 3636 is a perfect square (62=366^2 = 36):

x+5=±6x + 5 = \pm 6

Write as two equations and solve each:

x+5=6    x=1x + 5 = 6 \implies x = 1

x+5=6    x=11x + 5 = -6 \implies x = -11

Step 6 — Check both solutions:

For x=1x = 1: 12+10(1)+4=1+10+4=151^2 + 10(1) + 4 = 1 + 10 + 4 = 15

For x=11x = -11: (11)2+10(11)+4=121110+4=15(-11)^2 + 10(-11) + 4 = 121 - 110 + 4 = 15

The solutions are x=1x = 1 and x=11x = -11. Unlike earlier examples where the equation was already in the form x2+bx=cx^2 + bx = c, this equation has a constant on the same side as the variable terms. Step 1 requires subtracting that constant from both sides before the completing-the-square process can begin. The trinomial x2+10x+25x^2 + 10x + 25 matches the Binomial Squares Pattern because (x+5)2=x2+10x+25(x + 5)^2 = x^2 + 10x + 25.

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Updated 2026-04-21

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