Standard High contrast

Algebra

Question 4 of 6

Completing the Square

Question

Solve \(9y^2 - 24y + 13 = 0\) by “completing the square” (leave your answers in surd form)

Show solution Hide solution Fully worked — 8 steps
  1. Divide both sides by 9 (step 1)
    \[ 9y^2 - 24y + 13 = 0 \]
    \[ y^2 - \frac{8}{3}y + \frac{13}{9} = 0 \]
  2. Subtract \(\frac{13}{9}\) from both sides (step 2)
    \[ y^2 - \frac{8}{3}y = -\frac{13}{9} \]
  3. Take ½ the coefficient of \(y\) and square it (step 3)
    \[ \left(\frac{1}{2} \times \frac{8}{3}\right)^{2} = \frac{16}{9} \]
  4. Add the squared value to both sides (step 4)
    \[ y^2 - \frac{8}{3}y + \frac{16}{9} = -\frac{13}{9} + \frac{16}{9} \]
    \[ y^2 - \frac{8}{3}y + \frac{16}{9} = \frac{3}{9} \]
  5. Factor the trinomial (step 5)
    \[ \left(y - \frac{4}{3}\right)^{2} = \frac{3}{9} \]
  6. Take the square root of both sides (step 6)
    \[ y - \frac{4}{3} = \pm \frac{\sqrt{3}}{3} \]
  7. Solve for \(y\) (step 7): add \(\frac{4}{3}\) to both sides
    \[ y = \frac{4}{3} \pm \frac{\sqrt{3}}{3} \]
    \[ y = \frac{4 \pm \sqrt{3}}{3} \]
  8. Final answer

    So the roots are either \(y = \frac{4 + \sqrt{3}}{3}\) or \(y = \frac{4 - \sqrt{3}}{3}\). That is, \(y\) is four plus the square root of three, all over three; or \(y\) is four minus the square root of three, all over three.

    \[ y = \frac{4 + \sqrt{3}}{3} \quad \text{or} \quad y = \frac{4 - \sqrt{3}}{3} \]