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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Solving Square Root and Other Radical Equations

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Exercise 43 Page 395

Rearrange the radical equation so that one of the radical expressions is isolated. Then, raise both sides of the equation to a power equal to the index of the radicals.

$5$

Practice makes perfect

We will find and check the solutions of the given equation.

Finding the Solutions

To solve equations with a variable expression inside a radical, we will first rearrange the radical equation so that one of the radical expressions is isolated. Then, we can raise both sides of the equation to a power equal to the index of the radicals. Let's try to solve our equation using this method!

2 x + 6 - x - 1 = 2

$LHS + x - 1 = RHS + x - 1$

2 x + 6 = 2 + x - 1

$LHS^{2} = RHS^{2}$

(2 x + 6)^{2} = (2 + x - 1)^{2}

$(a)^{2} = a$

2 x + 6 = (2 + x - 1)^{2}

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Simplify right-hand side

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

2 x + 6 = 2^{2} + 2 (2) (x - 1) + (x - 1)^{2}

Calculate power and product

2 x + 6 = 4 + 4 x - 1 + (x - 1)^{2}

$(a)^{2} = a$

2 x + 6 = 4 + 4 x - 1 + x - 1

Subtract term

2 x + 6 = 3 + 4 x - 1 + x

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Subchapter links

Lesson Check p.394

Practice and Problem-Solving Exercises p.395-397

Standardized Test Prep p.397