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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 41 Page 395

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!

5 - x - x = 1

$LHS + x = RHS + x$

5 - x = x + 1

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

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

$(a)^{2} = a$

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

▼

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

ExpandPosPerfectSquare

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

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

$(a)^{2} = a$

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

Multiply

5 - x = x + 2 x + 1^{2}

$1^{a} = 1$

5 - x = x + 2 x + 1

Since we still have a radical expression on one side of the equation, we will isolate the radical and then follow the same method.

Subchapter links

Lesson Check p.394

Practice and Problem-Solving Exercises p.395-397

Standardized Test Prep p.397