Answers: Completing the square
These are the answers to Questions: Completing the square.
Please attempt the questions before reading these answers!
Q1
1.1. Here, \(x^2 - 2x + 15 = (x-1)^2 + 14\), so in this question \(p = -1\) and \(q = 14\).
1.2. Here, \(y^2 - 6y + 8 = (y-3)^2 - 1\), so in this question \(p = -3\) and \(q = -1\).
1.3. Here, \(x^2 + 8x + 20 = (x+4)^2 + 4\), so in this question \(p = 4\) and \(q = 4\).
1.4. Here, \(m^2 - 26m + 25 = (m-13)^2 - 144\), so in this question \(p = -13\) and \(q = -144\).
1.5. Here, \(n^2 + 6n + 50 = (m+3)^2 + 41\), so in this question \(p = 3\) and \(q = 41\).
1.6. Here, \(x^2 + 2x + 144 = (x+1)^2 + 143\), so in this question \(p = 1\) and \(q = 143\).
1.7. Here, \(\displaystyle h^2 - 3h - 3 = \left(h- \frac{3}{2}\right)^2 + \frac{3}{4}\), so in this question \(p = -3/2\) and \(q = 3/4\).
1.8. Here, \(\displaystyle x^2 + x - 3 = \left(x+ \frac{1}{2}\right)^2 - \frac{13}{4}\), so in this question \(p = 1/2\) and \(q = -13/4\).
1.9. Here, \(\displaystyle x^2 - 13x + 43 = \left(x- \frac{13}{2}\right)^2 + \frac{3}{4}\), so in this question \(p = -13/2\) and \(q = 3/4\).
1.10. Here, \(\displaystyle y^2 - 8y + 16 = \left(y- 4\right)^2\), so in this question \(p = -4\) and \(q = 0\).
1.11. Here, \(\displaystyle x^2 + 13x + 9 = \left(x+ \frac{13}{2}\right)^2 - \frac{133}{4}\), so in this question \(p = 13/2\) and \(q = -133/4\).
1.12. Here, \(\displaystyle m^2 + 3m + 33 = \left(m+ \frac{3}{2}\right)^2 - \frac{143}{4}\), so in this question \(p = 3/2\) and \(q = -143/4\).
Q2
2.1. Here, \(2x^2 - 12x + 14 = 2(x-3)^2 - 4\), so in this question \(a = 2\), \(p = -3\) and \(q = -4\).
2.2. Here, \(5y^2 - 10y + 4 = 5(x-1)^2 - 1\), so in this question \(a = 5\), \(p = -1\) and \(q = -1\).
2.3. Here, \(4x^2 + 32x + 68 = 4(x+4)^2 + 4\), so in this question \(a = p = q = 4\). (Or, if you prefer, \((2x+8)^2 + 4\).)
2.4. Here, \(\displaystyle 2m^2 + 2m + 2 = 2\left(m + \frac{1}{2}\right)^2 + \frac{3}{2}\), so in this question \(a = 2\), \(p = 1/2\) and \(q = 3/2\).
2.5. Here, \(\displaystyle 3x^2 - 2x + 5 = 3\left(x - \frac{1}{3}\right)^2 + \frac{14}{3}\), so in this question \(a = 3\), \(p = -1/3\) and \(q = 14/3\).
2.6. Here, \(\displaystyle 4x^2 - 4x + 1 = 4\left(x - \frac{1}{2}\right)^2\), so in this question \(a = 4\), \(p = -1/2\) and \(q = 0\). (Or, if you prefer, \((2x-1)^2\).)
2.7. Here, \(\displaystyle 2h^2 - 3h + 1 = 2\left(h- \frac{3}{4}\right)^2 - \frac{1}{8}\), so in this question \(a = 2\), \(p = -3/4\) and \(q = -1/8\).
2.8. Here, \(\displaystyle 3x^2 + 5x + 2 = 3\left(x+ \frac{5}{6}\right)^2 - \frac{3}{36}\), so in this question \(a = 3\), \(p = 5/6\) and \(q = -3/36\).
Q3
Using your working from Q1 and Q2, solve the following quadratic equations.
3.1. You worked out in 1.2 that \(y^2 - 6y + 8 = (y-3)^2 - 1\). Rearranging \((y-3)^2 - 1 = 0\) for \(y\) gives \(y = 3 \pm 1\), so \(y = 2\) or \(y = 4\).
3.2. You worked out in 1.4 that \(m^2 - 26m + 25 = (m-13)^2 - 144\). Rearranging \((y-3)^2 - 144 = 0\) for \(y\) gives \(y = 13 \pm 12\), so \(y = 1\) or \(y = 25\).
3.3. You worked out in 1.3 that \(x^2 + 8x + 20 = (x+4)^2 + 4\). Using the fact that \((\pm 2i)^2 = -4\) (see [Guide: Introduction to complex numbers]), rearranging \((x+4)^2 + 4 = 0\) for \(y\) gives \(y = -4 \pm 2i\), so \(y = -4 - 2i\) or \(y = -4 + 2i\).
3.4. You worked out in 2.6 that \(\displaystyle 4x^2 - 4x + 1 = 4\left(x - \frac{1}{2}\right)^2\). Rearranging \(4\left(x - \frac{1}{2}\right)^2 = 0\) for \(x\) gives \(\displaystyle x = \frac{1}{2}\) (twice, see Guide: Introduction to quadratic equations).
3.5. You worked out in 2.3 that \(4x^2 + 32x + 68 = 4(x+4)^2 + 4\). Using the fact that \((\pm i)^2 = -1\) (see [Guide: Introduction to complex numbers]), rearranging \(4(x+4)^2 + 4 = 0\) for \(x\) gives \(x = -4 \pm i\), so \(x = -4 - i\) or \(x = -4 + i\).
3.6. You worked out in 2.8 that \(\displaystyle 3x^2 + 5x + 2 = 3\left(x+ \frac{5}{6}\right)^2 - \frac{3}{36}\). Rearranging \(\displaystyle 3\left(x+ \frac{5}{6}\right)^2 - \frac{3}{36} = 0\) for \(x\) gives \(\displaystyle y = -\frac{5}{6} \pm \frac{1}{6}\), so \(y = -1\) or \(y = -2/3\).
Version history and licensing
v1.0: initial version created 09/24 by tdhc.