Example

Solving 3x2−12x−15=03x^2 - 12x - 15 = 0 by Completing the Square

Solve 3x2−12x−15=03x^2 - 12x - 15 = 0 by completing the square, demonstrating the procedure when the leading coefficient is not 11 but can be factored out as a GCF. Preliminary step — Make the leading coefficient 11. The coefficient of x2x^2 is 33, which divides evenly into all three terms. Factor out 33: 3(x2−4x−5)=03(x^2 - 4x - 5) = 0 Divide both sides by 33: x2−4x−5=0x^2 - 4x - 5 = 0 Step 1 — Isolate the variable terms. Add 55 to both sides to move the constant to the right: x2−4x=5x^2 - 4x = 5 Step 2 — Find (12⋅b)2\left(\frac{1}{2} \cdot b\right)^2 and add it to both sides. The coefficient of xx is −4-4, so b=−4b = -4. Compute: (12(−4))2=(−2)2=4\left(\frac{1}{2}(-4)\right)^2 = (-2)^2 = 4. Add 44 to both sides: x2−4x+4=5+4x^2 - 4x + 4 = 5 + 4 Step 3 — Factor the perfect square trinomial. The left side factors as a binomial square using the subtraction form: (x−2)2=9(x - 2)^2 = 9 Step 4 — Apply the Square Root Property: x−2=±9x - 2 = \pm\sqrt{9} Step 5 — Simplify and solve. Since 99 is a perfect square (32=93^2 = 9): x−2=±3x - 2 = \pm 3 Write as two equations and solve each: x−2=3  ⟹  x=5x - 2 = 3 \implies x = 5 x−2=−3  ⟹  x=−1x - 2 = -3 \implies x = -1 Step 6 — Check both solutions: For x=5x = 5: 3(5)2−12(5)−15=75−60−15=03(5)^2 - 12(5) - 15 = 75 - 60 - 15 = 0 ✓ For x=−1x = -1: 3(−1)2−12(−1)−15=3+12−15=03(-1)^2 - 12(-1) - 15 = 3 + 12 - 15 = 0 ✓ The solutions are x=5x = 5 and x=−1x = -1. This example introduces a preliminary step not present in earlier completing-the-square problems: the leading coefficient 33 must be removed before the standard procedure can begin. Because 33 divides evenly into all three coefficients (33, −12-12, and −15-15), factoring it out and dividing both sides by 33 reduces the equation to the familiar form x2+bx+c=0x^2 + bx + c = 0 with a leading coefficient of 11.

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Updated 2026-06-30

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