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Algebra

Question 2 of 4

The Quadratic Formula

Question

Find the solution of \(12y^2 + 4y = 1\) using the quadratic formula

Show solution Hide solution Fully worked — 9 steps
  1. Check if it is in the form \(ax^2 + bx + c = 0\)
    \[ 12y^2 + 4y = 1 \]
  2. Rearrange in the form \(ax^2 + bx + c = 0\)
    \[ 12y^2 + 4y - 1 = 0 \]
  3. Identify the values of \(a\), \(b\) and \(c\)
    \[ a = 12 \qquad b = 4 \qquad c = -1 \]
  4. Use the formula to find the solutions
    \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  5. Substitute the values of \(a\), \(b\) and \(c\) and solve
    \[ y = \frac{-(4) \pm \sqrt{(4)^2 - 4(12)(-1)}}{2(12)} \]
  6. Evaluate each part

    \(-(4) = -4\); \((4)^2 = 16\); \(-4(12)(-1) = +48\) and \(2(12) = 24\):

    \[ = \frac{-4 \pm \sqrt{16 + 48}}{24} \]
  7. Simplify under the square root

    \(16 + 48 = 64\):

    \[ = \frac{-4 \pm \sqrt{64}}{24} \]

    \(\sqrt{64} = 8\):

    \[ = \frac{-4 \pm 8}{24} \]
  8. Find the two solutions

    Then

    \[ y = \frac{-4 + 8}{24} = \frac{1}{6} \]

    or

    \[ y = \frac{-4 - 8}{24} = -\frac{1}{2} \]
  9. State the solutions

    So the solutions are \(y = \frac{1}{6}\) and \(y = -\frac{1}{2}\). That is, \(y\) is one sixth and \(y\) is minus one half.