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Example

Solving n2=3n+11n^2 = 3n + 11 by Completing the Square

Solve n2=3n+11n^2 = 3n + 11 by completing the square, demonstrating the procedure when variable terms must first be collected on one side and the linear coefficient is an odd number.

Step 1 — Collect the variable terms on one side. Subtract 3n3n from both sides to move all variable terms to the left:

n23n=11n^2 - 3n = 11

Step 2 — Find (12b)2\left(\frac{1}{2} \cdot b\right)^2 and add it to both sides. The coefficient of nn is 3-3, so b=3b = -3. Compute: (12(3))2=(32)2=94\left(\frac{1}{2}(-3)\right)^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4}. Add 94\frac{9}{4} to both sides:

n23n+94=11+94n^2 - 3n + \frac{9}{4} = 11 + \frac{9}{4}

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

(n32)2=444+94=534\left(n - \frac{3}{2}\right)^2 = \frac{44}{4} + \frac{9}{4} = \frac{53}{4}

Step 4 — Apply the Square Root Property:

n32=±534n - \frac{3}{2} = \pm\sqrt{\frac{53}{4}}

Step 5 — Simplify the radical and solve. Apply the Quotient Property to the radical: 534=534=532\sqrt{\frac{53}{4}} = \frac{\sqrt{53}}{\sqrt{4}} = \frac{\sqrt{53}}{2}. Since 5353 is prime, 53\sqrt{53} cannot be simplified further. Add 32\frac{3}{2} to both sides:

n=32±532n = \frac{3}{2} \pm \frac{\sqrt{53}}{2}

Write as two solutions:

n=32+532orn=32532n = \frac{3}{2} + \frac{\sqrt{53}}{2} \quad \text{or} \quad n = \frac{3}{2} - \frac{\sqrt{53}}{2}

The solutions are n=3+532n = \frac{3 + \sqrt{53}}{2} and n=3532n = \frac{3 - \sqrt{53}}{2}. This example combines two complications: first, the variable terms appear on both sides of the original equation, so they must be collected to one side before the procedure can begin. Second, the linear coefficient 3-3 is odd, so halving it produces the fraction 32-\frac{3}{2}, and the resulting constant 94\frac{9}{4} introduces fractions throughout the remaining steps. The radicand 5353 is not a perfect square, so the final solutions remain in radical form.

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Updated 2026-04-21

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