1Cademy - Solving $$2x^2 - 3x = 20$$ by Completing the Square

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Completing the Square When the Leading Coefficient Is Not One

Example

Solving $2x^2 - 3x = 20$ by Completing the Square

Solve $2x^2 - 3x = 20$ by completing the square, demonstrating Strategy 2 for handling a leading coefficient that is not $1$ — dividing both sides by the leading coefficient to produce fraction coefficients.

Preliminary step — Make the leading coefficient $1$ . The coefficient of $x^2$ is $2$ . Because $2$ does not divide evenly into the linear coefficient $-3$ , divide every term on both sides by $2$ :

$x^2 - \frac{3}{2}x = 10$

Step 1 — Isolate the variable terms. The variable terms are already on the left and the constant is on the right.

Step 2 — Find $\left(\frac{1}{2} \cdot b\right)^2$ and add it to both sides. The coefficient of $x$ is $-\frac{3}{2}$ , so $b = -\frac{3}{2}$ . Take half of $-\frac{3}{2}$ : $\frac{1}{2}\left(-\frac{3}{2}\right) = -\frac{3}{4}$ . Square the result: $\left(-\frac{3}{4}\right)^2 = \frac{9}{16}$ . Add $\frac{9}{16}$ to both sides:

$x^2 - \frac{3}{2}x + \frac{9}{16} = 10 + \frac{9}{16}$

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

$\left(x - \frac{3}{4}\right)^2 = \frac{160}{16} + \frac{9}{16} = \frac{169}{16}$

Step 4 — Apply the Square Root Property:

$x - \frac{3}{4} = \pm\sqrt{\frac{169}{16}}$

Step 5 — Simplify the radical and solve. Since both $169 = 13^2$ and $16 = 4^2$ are perfect squares:

$x - \frac{3}{4} = \pm\frac{13}{4}$

Solve for $x$ by adding $\frac{3}{4}$ to both sides:

$x = \frac{3}{4} + \frac{13}{4} = \frac{16}{4} = 4 \qquad \text{or} \qquad x = \frac{3}{4} - \frac{13}{4} = \frac{-10}{4} = -\frac{5}{2}$

The solutions are $x = 4$ and $x = -\frac{5}{2}$ . This example illustrates Strategy 2 for completing the square when the leading coefficient is not $1$ : dividing both sides by the leading coefficient introduces fraction coefficients throughout the problem. Despite the fractions, the right side $\frac{169}{16}$ turns out to be a perfect square fraction, so the final solutions are rational numbers.

Updated 2026-04-21

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OpenStax Elementary Algebra

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