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Solve square-root equations with real and complex answers

Keep both signs of the square root

From $u^{2} = c$ , write $u = \pm c$ . If $c < 0$ , the solutions are complex because $- d = i d$ for positive $d$ . If only real solutions are requested, report none for that case.

Problem

Solve over the complex numbers: (9) $r^{2} = - 44$ ; (10) $r^{2} = 48$ ; (11) $2 b^{2} + 1 = - 2$ ; (12) $2 x^{2} - 10 = 38$ ; (13) $(x + 3)^{2} = 20$ ; (14) $(x - 5)^{2} = - 36$ .

Answer

(9)

r = \pm 2 i 11

; (10)

r = \pm 43

; (11)

b = \pm i 6 /2

; (12)

x = \pm 26

; (13)

x = - 3 \pm 25

; (14)

x = 5 \pm 6 i

Problems 9, 11 and 14 have no real solutions; their answers here are complex.

Step-by-step solution

1. Isolate each square

The equations become $r^{2} = - 44$ , $r^{2} = 48$ , $b^{2} = - 3/2$ , $x^{2} = 24$ , $(x + 3)^{2} = 20$ and $(x - 5)^{2} = - 36$ .

2. Take both signs

Use $u = \pm c$ . Simplifying radicals gives $r = \pm 2 i 11$ , $r = \pm 43$ , $b = \pm i 6 /2$ and $x = \pm 26$ for the first four.

3. Undo the shifts

From $x + 3 = \pm 25$ , subtract 3: $x = - 3 \pm 25$ . From $x - 5 = \pm 6 i$ , add 5: $x = 5 \pm 6 i$ .

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