Answers: Using the quadratic formula
These are the answers to Questions: Using the quadratic formula.
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Answers
Q1
1.1. The two roots of \(x^2 - 7x + 6 = 0\) are \(x = 1\) and \(x = 6\).
1.2. The two roots of \(x^2 + 14x + 45 = 0\) are \(x=-9\) and \(x=-5\).
1.3. The two roots of \(x^2 - 4x + 13 = 0\) are \(x=2-3i\) and \(x=2+3i\).
1.4. The two roots of \(x^2 - x - 56 = 0\) are \(x = -7\) and \(x=8\).
1.5. The one distinct root of \(s^2 + 4s + 4 = 0\) is \(x = -2\).
1.6. The two roots of \(t^2 + 4t - 4 = 0\) are \(t = -2 - 2\sqrt{2}\) and \(t = -2 + 2\sqrt{2}\)
1.7. The two roots of \(m^2 - 144 = 0\) are \(m = -12\) and \(m = 12\).
1.8. The two roots of \(5c^2 - 25 + 30 = 0\) are \(c = -1\) and \(c=1\).
1.9. The two roots of \(2n^2 + n + 1 = 0\) are \(\displaystyle n = \frac{-1 - i\sqrt{7}}{4}\) and \(\displaystyle n = \frac{-1 + i\sqrt{7}}{4}\)
1.10. The two roots of \(-3c^2 + 9c - 1 = 0\) are \(c = \frac{3}{2} - \frac{\sqrt{69}}{6}\) and \(c = \frac{3}{2} + \frac{\sqrt{69}}{6}\).
1.11. The two roots of \(\frac{x^2}{2} - \frac{7x}{2} + 3 = 0\) are \(x = 1\) and \(x = 6\).
1.12. The one distinct root of \(e^{2x} - 4e^x + 4 = 0\) is \(e^x = 2\), giving \(x = \ln(2)\) as a solution.
1.13. The two roots of \(-9s^2 + 3s - 1 = 0\) are \(\displaystyle s = \frac{1 - i\sqrt{3}}{6}\) and \(\displaystyle s = \frac{1 + i\sqrt{3}}{6}\).
1.14. The two roots of \(2e^{6x} + e^{3x} + 1 = 0\) are \(\displaystyle e^{3x} = \frac{-1 - i\sqrt{7}}{4}\) and \(\displaystyle e^{3x} = \frac{-1 + i\sqrt{7}}{4}\), and so there are no real solutions for \(x\).
1.15. The one distinct root of \(\cos^2(x) + 4\cos(x) - 4 = 0\) is \(\cos(x) = 2\), and so there are no real solutions for \(x\) as \(-1\leq \cos(x) \leq 1\) for all real \(x\).
1.16. The two distinct roots of \(8m^2 - 4m - 1 = 0\) are \(\displaystyle m = \frac{1-\sqrt{3}}{4}\) and \(\displaystyle m = \frac{1+\sqrt{3}}{4}\)
Q2
In Questions: Introduction to quadratic equations, you saw that the following expressions are all quadratic equations in disguise. Solve these for the variable indicated.
2.1. The two roots of \(x = 1/x - 1\) are \(\displaystyle x = \frac{-1-\sqrt{5}}{2}\) and \(\displaystyle x = \frac{-1+\sqrt{5}}{2}\).
2.2. The two roots of \((y-1)(y-4) = -(y+2)(y+3)\) are \(y = -i\sqrt{5}\) and \(y = i\sqrt{5}\).
2.3. The one distinct root of \(4m(m+1) + 6 = 5\) is \(m = -1/2\).
2.4. The two roots of \((t-1)(t+1) = -2\) are \(t = -i\) and \(t=i\).
2.5. The two roots of \(\frac{x-1}{x-2} = 5x\) are \(\displaystyle x=\frac{11-\sqrt{101}}{10}\) and \(\displaystyle x=\frac{11+\sqrt{101}}{10}\).
2.6. The two solutions in \(e^x\) for \(\frac{e^{x} - e^{-x}}{2} = 1\) are \(e^x = 1 - \sqrt{2}\) and \(e^x = 1 + \sqrt{2}\). Of these, \(x = \ln(1+\sqrt{2})\) is a valid solution in \(x\), as \(e^x\) cannot be negative.
Version history and licensing
v1.0: initial version created 04/23 by tdhc.
- v1.1: edited 05/24 by tdhc.