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Algebra

Question 5 of 6

Completing the Square

Question

Solve \(x^2 - x = \frac{1}{2}\) by “completing the square” (leave your answers in surd form)

Show solution Hide solution Fully worked — 7 steps
  1. Check the arrangement of the equation

    Since both terms containing variables are on one side and the constant is on the other, proceed to step 3.

    \[ x^2 - x = \frac{1}{2} \]
  2. Take ½ the coefficient of \(x\) and square it (step 3)
    \[ \left(\frac{1}{2} \times 1\right)^{2} = \frac{1}{4} \]
  3. Add the squared value to both sides (step 4)
    \[ x^2 - x + \frac{1}{4} = \frac{1}{2} + \frac{1}{4} \]
    \[ x^2 - x + \frac{1}{4} = \frac{3}{4} \]
  4. Factor the trinomial (step 5)
    \[ \left(x - \frac{1}{2}\right)^{2} = \frac{3}{4} \]
  5. Take the square root of both sides (step 6)
    \[ x - \frac{1}{2} = \pm \frac{\sqrt{3}}{2} \]
  6. Solve for \(x\) (step 7): add \(\frac{1}{2}\) to both sides
    \[ x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} \]
    \[ x = \frac{1 \pm \sqrt{3}}{2} \]
  7. Final answer

    So the roots are either \(x = \frac{1 + \sqrt{3}}{2}\) or \(x = \frac{1 - \sqrt{3}}{2}\). That is, \(x\) is one plus the square root of three, all over two; or \(x\) is one minus the square root of three, all over two.

    \[ x = \frac{1 + \sqrt{3}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{3}}{2} \]