body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Pearson Algebra 2 Common Core, 2011

PA

Pearson Algebra 2 Common Core, 2011 View details

5. Solving Square Root and Other Radical Equations

Continue to next subchapter

Exercise 41 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.

$1$

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!

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