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Algebra

Question 6 of 6

Completing the Square

Question

Solve \(2z^2 - 7z = 15\) by “completing the square” (leave your answers in surd form)

Show solution Hide solution Fully worked — 7 steps
  1. Divide both sides by 2 (step 1)
    \[ 2z^2 - 7z = 15 \]
    \[ z^2 - \frac{7}{2}z = \frac{15}{2} \]
  2. Take ½ the coefficient of \(z\) and square it (step 3)
    \[ \left(\frac{1}{2} \times \frac{7}{2}\right)^{2} = \frac{49}{16} \]
  3. Add the squared value to both sides (step 4)
    \[ z^2 - \frac{7}{2}z + \frac{49}{16} = \frac{15}{2} + \frac{49}{16} \]
    \[ z^2 - \frac{7}{2}z + \frac{49}{16} = \frac{169}{16} \]
  4. Factor the trinomial (step 5)
    \[ \left(z - \frac{7}{4}\right)^{2} = \frac{169}{16} \]
  5. Take the square root of both sides (step 6)
    \[ z - \frac{7}{4} = \pm \frac{13}{4} \]
  6. Solve for \(z\) (step 7): add \(\frac{7}{4}\) to both sides
    \[ z = \frac{7}{4} \pm \frac{13}{4} \]
    \[ z = \frac{7 \pm 13}{4} \]
    \[ z = \frac{7 + 13}{4} \quad \text{or} \quad z = \frac{7 - 13}{4} \]
  7. Final answer

    So the roots are either \(z = 5\) or \(z = -\frac{3}{2}\). That is, \(z\) is five, or \(z\) is minus three over two.

    \[ z = 5 \quad \text{or} \quad z = -\frac{3}{2} \]