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## Solve by Extracting Square Roots

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## Unit 14: Lesson 3

Solving quadratics by taking square roots.

- Solving quadratics by taking square roots examples
- Quadratics by taking square roots: strategy
- Solving quadratics by taking square roots: with steps
- Solving simple quadratics review

## What you should be familiar with before taking this lesson

## What you will learn in this lesson

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 Â 3 / 4 1\ 3/4 1 Â 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 Â pi 12\ \text{pi} 1 2 Â pi 12, space, start text, p, i, end text or 2 / 3 Â pi 2/3\ \text{pi} 2 / 3 Â pi 2, slash, 3, space, start text, p, i, end text
- (Choice A) Â x = Â± 5 x=\pm 5 x = Â± 5 x, equals, plus minus, 5 A x = Â± 5 x=\pm 5 x = Â± 5 x, equals, plus minus, 5
- (Choice B) Â x = Â± 5 x=\pm\sqrt{5} x = Â± 5 â€‹ x, equals, plus minus, square root of, 5, end square root B x = Â± 5 x=\pm\sqrt{5} x = Â± 5 â€‹ x, equals, plus minus, square root of, 5, end square root
- (Choice C) Â x = Â± 25 x=\pm 25 x = Â± 2 5 x, equals, plus minus, 25 C x = Â± 25 x=\pm 25 x = Â± 2 5 x, equals, plus minus, 25

## Solving ( x âˆ’ 2 ) 2 = 49 (x-2)^2=49 ( x âˆ’ 2 ) 2 = 4 9 left parenthesis, x, minus, 2, right parenthesis, squared, equals, 49 and similar equations

Isolating x x x x, understanding the solutions.

- (Choice A) Â x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2 and x = 8 x=8 x = 8 x, equals, 8 A x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2 and x = 8 x=8 x = 8 x, equals, 8
- (Choice B) Â x = Â± 2 x=\pm 2 x = Â± 2 x, equals, plus minus, 2 B x = Â± 2 x=\pm 2 x = Â± 2 x, equals, plus minus, 2
- (Choice C) Â x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 8 x=-8 x = âˆ’ 8 x, equals, minus, 8 C x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 8 x=-8 x = âˆ’ 8 x, equals, minus, 8
- (Choice D) Â x = Â± 5 x=\pm 5 x = Â± 5 x, equals, plus minus, 5 D x = Â± 5 x=\pm 5 x = Â± 5 x, equals, plus minus, 5
- (Choice A) Â x = Â± 3 x=\pm 3 x = Â± 3 x, equals, plus minus, 3 A x = Â± 3 x=\pm 3 x = Â± 3 x, equals, plus minus, 3
- (Choice B) Â x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1 B x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1
- (Choice C) Â x = 4 x=4 x = 4 x, equals, 4 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2 C x = 4 x=4 x = 4 x, equals, 4 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2
- (Choice D) Â x = 5 x=5 x = 5 x, equals, 5 and x = âˆ’ 4 x=-4 x = âˆ’ 4 x, equals, minus, 4 D x = 5 x=5 x = 5 x, equals, 5 and x = âˆ’ 4 x=-4 x = âˆ’ 4 x, equals, minus, 4
- (Choice A) Â x = 7 + 5 x=\sqrt{7}+5 x = 7 â€‹ + 5 x, equals, square root of, 7, end square root, plus, 5 and x = âˆ’ 7 + 5 x=-\sqrt{7}+5 x = âˆ’ 7 â€‹ + 5 x, equals, minus, square root of, 7, end square root, plus, 5 A x = 7 + 5 x=\sqrt{7}+5 x = 7 â€‹ + 5 x, equals, square root of, 7, end square root, plus, 5 and x = âˆ’ 7 + 5 x=-\sqrt{7}+5 x = âˆ’ 7 â€‹ + 5 x, equals, minus, square root of, 7, end square root, plus, 5
- (Choice B) Â x = 7 + 5 x=\sqrt{7}+5 x = 7 â€‹ + 5 x, equals, square root of, 7, end square root, plus, 5 and x = 7 âˆ’ 5 x=\sqrt{7}-5 x = 7 â€‹ âˆ’ 5 x, equals, square root of, 7, end square root, minus, 5 B x = 7 + 5 x=\sqrt{7}+5 x = 7 â€‹ + 5 x, equals, square root of, 7, end square root, plus, 5 and x = 7 âˆ’ 5 x=\sqrt{7}-5 x = 7 â€‹ âˆ’ 5 x, equals, square root of, 7, end square root, minus, 5
- (Choice C) Â x = 5 + 7 x=\sqrt{5}+7 x = 5 â€‹ + 7 x, equals, square root of, 5, end square root, plus, 7 and x = âˆ’ 5 + 7 x=-\sqrt{5}+7 x = âˆ’ 5 â€‹ + 7 x, equals, minus, square root of, 5, end square root, plus, 7 C x = 5 + 7 x=\sqrt{5}+7 x = 5 â€‹ + 7 x, equals, square root of, 5, end square root, plus, 7 and x = âˆ’ 5 + 7 x=-\sqrt{5}+7 x = âˆ’ 5 â€‹ + 7 x, equals, minus, square root of, 5, end square root, plus, 7
- (Choice D) Â x = 12 x=12 x = 1 2 x, equals, 12 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2 D x = 12 x=12 x = 1 2 x, equals, 12 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2

## Why we shouldn't expand the parentheses

- (Choice A) Â x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 6 x=-6 x = âˆ’ 6 x, equals, minus, 6 A x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 6 x=-6 x = âˆ’ 6 x, equals, minus, 6
- (Choice B) Â x = Â± 6 x=\pm 6 x = Â± 6 x, equals, plus minus, 6 B x = Â± 6 x=\pm 6 x = Â± 6 x, equals, plus minus, 6
- (Choice C) Â x = 6 x=6 x = 6 x, equals, 6 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2 C x = 6 x=6 x = 6 x, equals, 6 and x = âˆ’ 2 x=-2 x = âˆ’ 2 x, equals, minus, 2
- (Choice D) Â x = Â± 2 x=\pm 2 x = Â± 2 x, equals, plus minus, 2 D x = Â± 2 x=\pm 2 x = Â± 2 x, equals, plus minus, 2
- (Choice A) Â x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1 A x = 2 x=2 x = 2 x, equals, 2 and x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1
- (Choice B) Â x = Â± 3 x=\pm 3 x = Â± 3 x, equals, plus minus, 3 B x = Â± 3 x=\pm 3 x = Â± 3 x, equals, plus minus, 3
- (Choice D) Â x = Â± 6 x=\pm 6 x = Â± 6 x, equals, plus minus, 6 D x = Â± 6 x=\pm 6 x = Â± 6 x, equals, plus minus, 6
- (Choice A) Â x = 11 x=11 x = 1 1 x, equals, 11 and x = 5 x=5 x = 5 x, equals, 5 A x = 11 x=11 x = 1 1 x, equals, 11 and x = 5 x=5 x = 5 x, equals, 5
- (Choice B) Â x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1 and x = âˆ’ 7 x=-7 x = âˆ’ 7 x, equals, minus, 7 B x = âˆ’ 1 x=-1 x = âˆ’ 1 x, equals, minus, 1 and x = âˆ’ 7 x=-7 x = âˆ’ 7 x, equals, minus, 7

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## VIDEO

## COMMENTS

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the square root of both sides of the equation to solve for . This is known as Extracting Square Roots. 2 âˆ’ 25 = 0. Page 2

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