# Lesson 18

The Quadratic Formula and Complex Solutions

### Problem 1

Clare solves the quadratic equation \(4x^2+12x+58=0\), but when she checks her answer, she realizes she made a mistake. Explain what Clare's mistake was.

\(\begin{align} x &= \frac{\text-12 \pm \sqrt{12^2 - 4 \boldcdot 4 \boldcdot 58}}{2 \boldcdot 4} \\ x &= \frac{\text-12 \pm \sqrt{144-928}}{8} \\ x &= \frac{\text-12 \pm \sqrt{\text-784}}{8} \\ x &= \frac{\text-12 \pm 28i}{8} \\ x &= \text-1.5 \pm 28i \\ \end{align}\)

### Solution

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### Problem 2

Write in the form \(a+bi\), where \(a\) and \(b\) are real numbers:

- \(\frac{5 \pm \sqrt{\text-4}}{3}\)
- \(\frac{10 \pm \sqrt{\text-16}}{2}\)
- \(\frac{\text-3 \pm \sqrt{\text-144}}{6}\)

### Solution

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### Problem 3

Priya is using the quadratic formula to solve two different quadratic equations.

For the first equation, she writes \(x= \frac{4 \pm \sqrt{16-72}}{12}\)

For the second equation, she writes \(x=\frac{8 \pm \sqrt{64-24}}{6}\)

Which equation(s) will have real solutions? Which equation(s) will have non-real solutions? Explain how you know.

### Solution

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### Problem 4

Find the exact solution(s) to each of these equations, or explain why there is no solution.

- \(x^2=25\)
- \(x^3 = 27\)
- \(x^2=12\)
- \(x^3=12\)

### Solution

### Problem 5

Kiran is solving the equation \(\sqrt{x+2} - 5 = 11\) and decides to start by squaring both sides. Which equation results if Kiran squares both sides as his first step?

\(x + 2 - 25 = 121\)

\(x + 2 + 25 = 121\)

\(x+2 - 10\sqrt{x+2} + 25 = 121\)

\(x+2 + 10\sqrt{x+2} + 25 = 121\)

### Solution

### Problem 6

Plot each number on the real or imaginary number line.

- \(\text-\sqrt{4}\)
- \(\sqrt{\text-1}\)
- \(3\sqrt{4}\)
- \(\text-3\sqrt{\text-1}\)
- \(4\sqrt{\text-1}\)
- \(2\sqrt{2}\)