1Cademy - Solving $$(x - 3)(x + 5) = 9$$ by Completing the Square

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Completing the Square

Example

Solving $(x - 3)(x + 5) = 9$ by Completing the Square

Solve $(x - 3)(x + 5) = 9$ by completing the square, demonstrating the procedure when the equation begins as a product of two distinct binomials equal to a nonzero constant.

Multiply the binomials on the left. Apply the FOIL method to expand $(x - 3)(x + 5)$ : First $x \cdot x = x^2$ , Outer $x \cdot 5 = 5x$ , Inner $(-3) \cdot x = -3x$ , Last $(-3) \cdot 5 = -15$ . Combine like terms:

$x^2 + 2x - 15 = 9$

Step 1 — Isolate the variable terms. Add $15$ to both sides to move the constant away from the variable terms:

$x^2 + 2x = 24$

Step 2 — Find $\left(\frac{1}{2} \cdot b\right)^2$ and add it to both sides. The coefficient of $x$ is $2$ , so $b = 2$ . Compute: $\left(\frac{1}{2}(2)\right)^2 = 1^2 = 1$ . Add $1$ to both sides:

$x^2 + 2x + 1 = 24 + 1$

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

$(x + 1)^2 = 25$

Step 4 — Apply the Square Root Property:

$x + 1 = \pm\sqrt{25}$

Step 5 — Simplify and solve. Since $25$ is a perfect square ( $5^2 = 25$ ):

$x + 1 = \pm 5$

Write as two equations and solve each:

$x + 1 = 5 \implies x = 4$

$x + 1 = -5 \implies x = -6$

The solutions are $x = 4$ and $x = -6$ . This example introduces an additional preliminary step not present in earlier completing-the-square problems: the equation is given as a product of two different binomials equal to a constant, so the binomials must first be multiplied out using FOIL before the standard six-step procedure can begin. The expanded form $x^2 + 2x - 15 = 9$ then has constants on both sides, requiring the constant $-15$ to be moved to the right before completing the square.

Updated 2026-04-21

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OpenStax Elementary Algebra

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