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.

McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

4. Radical Equations

Continue to next subchapter

Exercise 5 Page 644

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 first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

3 x - 5 = x - 5

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

(3 x - 5)^{2} = (x - 5)^{2}

$(a)^{2} = a$

3 x - 5 = (x - 5)^{2}

ExpandNegPerfectSquare

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

3 x - 5 = x^{2} - 10 x + 25

▼

LHS - (3 x - 5) = RHS - (3 x - 5)

$LHS - 3 x = RHS - 3 x$

- 5 = x^{2} - 13 x + 25

$LHS + 5 = RHS + 5$

0 = x^{2} - 13 x + 30

Rearrange equation

x^{2} - 13 x + 30 = 0

We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of

a,

b,

and

c .

Subchapter links

Check Your Understanding p.644

H.O.T. Problems p.645

Standardized Test Practice p.646

Skills Review p.646