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Solving 3x2+2x=43x^2 + 2x = 4 by Completing the Square

Solve 3x2+2x=43x^2 + 2x = 4 by completing the square, demonstrating the procedure when dividing by the leading coefficient produces fractions and the resulting radicand is not a perfect square.

Preliminary step — Make the leading coefficient 11. The coefficient of x2x^2 is 33. Divide every term on both sides by 33:

x2+23x=43x^2 + \frac{2}{3}x = \frac{4}{3}

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

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 23\frac{2}{3}, so b=23b = \frac{2}{3}. Take half of 23\frac{2}{3}: 12⋅23=13\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}. Square the result: (13)2=19\left(\frac{1}{3}\right)^2 = \frac{1}{9}. Add 19\frac{1}{9} to both sides:

x2+23x+19=43+19x^2 + \frac{2}{3}x + \frac{1}{9} = \frac{4}{3} + \frac{1}{9}

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

(x+13)2=129+19=139\left(x + \frac{1}{3}\right)^2 = \frac{12}{9} + \frac{1}{9} = \frac{13}{9}

Step 4 — Apply the Square Root Property:

x+13=±139x + \frac{1}{3} = \pm\sqrt{\frac{13}{9}}

Step 5 — Simplify the radical and solve. Apply the Quotient Property: 139=139=133\sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{\sqrt{9}} = \frac{\sqrt{13}}{3}. Since 1313 is prime, 13\sqrt{13} cannot be simplified further. Subtract 13\frac{1}{3} from both sides:

x=−13±133x = -\frac{1}{3} \pm \frac{\sqrt{13}}{3}

Write as two solutions:

x=−13+133=−1+133orx=−13−133=−1−133x = -\frac{1}{3} + \frac{\sqrt{13}}{3} = \frac{-1 + \sqrt{13}}{3} \qquad \text{or} \qquad x = -\frac{1}{3} - \frac{\sqrt{13}}{3} = \frac{-1 - \sqrt{13}}{3}

The solutions are x=−1+133x = \frac{-1 + \sqrt{13}}{3} and x=−1−133x = \frac{-1 - \sqrt{13}}{3}. This example combines two complications: dividing by the leading coefficient 33 introduces fractions at every step, and the radicand 139\frac{13}{9} does not simplify to a perfect square, so the final answers remain in radical form. Unlike the previous example where 16916\frac{169}{16} was a perfect square fraction yielding rational solutions, here the prime radicand 1313 produces irrational solutions.

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Updated 2026-04-21

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References

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OpenStax

Elementary Algebra @ OpenStax

Ch.10 Quadratic Equations - Elementary Algebra @ OpenStax

Algebra

Math

Prealgebra

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