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.

Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

4. Solving Radical Equations and Inequalities

Continue to next subchapter

Exercise 55 Page 267

Practice makes perfect

a Let's solve the equations one at a time.

Solving the first equation

To solve the equation, we have to square both sides to eliminate the radicand.

x - 4 = 2 x

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

(x - 4)^{2} = 2 x

ExpandNegPerfectSquare

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

x^{2} - 8 x + 16 = 2 x

$LHS - 2 x = RHS - 2 x$

x^{2} - 10 x + 16 = 0

To solve this equation, we can use the Quadratic Formula.

x^{2} - 10 x + 16 = 0

Use the Quadratic Formula: $a = 1, b = - 10, c = 16$

x = \frac{- ( - 1 0 ) \pm ( - 1 0 ) ^{2} - 4 ( 1 ) ( 1 6 )}{2 \cdot 1}

▼

Simplify right-hand side

$- (- a) = a$

x = \frac{1 0 \pm ( - 1 0 ) ^{2} - 4 ( 1 ) ( 1 6 )}{2 \cdot 1}

Calculate power and product

x = \frac{1 0 \pm 1 0 0 - 6 4}{2}

Add terms

x = \frac{1 0 \pm 3 6}{2}

Calculate root

x = \frac{1 0 \pm 6}{2}

From here, we break the solution into two cases and solve each.

Positive : Negative : x_{1} = \frac{1 0 + 6}{2} \Leftrightarrow x_{1} = 8 x_{2} = \frac{1 0 - 6}{2} \Leftrightarrow x_{2} = 2

Squaring the equation, gives us a second degree equation with two solutions. Therefore, we have to investigate if any of them is extraneous by substituting them into the original equation and checking if the equation holds true.

x - 4 = 2 x

$x = 2$

2 - 4 = ? 2 (2)

▼

Simplify

Multiply

2 - 4 = ? 4

Calculate root

2 - 4 = ? 2

Subtract term

- 2 \neq = 2

The solution

x = 2

is extraneous. Let's check the other solution.

x - 4 = 2 x

$x = 8$

8 - 4 = ? 2 (8)

▼

Simplify

Multiply

8 - 4 = ? 16

Calculate root

8 - 4 = ? 4

Subtract term

4 = 4

The second solution

x = 8

is valid.

Solving the second equation

To solve the equation, we have to square both sides to eliminate the radicand. Note that the minus sign will disappear when you square the right-hand side.

x - 4 = - 2 x

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

(x - 4)^{2} = 2 x

ExpandNegPerfectSquare

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

x^{2} - 8 x + 16 = 2 x

$LHS - 2 x = RHS - 2 x$

x^{2} - 10 x + 16 = 0

Note that this is the same equation as we previously solved. Therefore, our solutions will still be

x = 2

and

x = 8 .

Like before, we have to test them in the original equation.

x - 4 = - 2 x

$x = 2$

2 - 4 = ? - 2 (2)

▼

Simplify

Multiply

2 - 4 = ? - 4

Calculate root

2 - 4 = ? - 2

Subtract term

- 2 = - 2

Now,

x = 2,

is a valid solution. Let's also check the second solution.

x - 4 = - 2 x

$x = 8$

8 - 4 = ? - 2 (8)

▼

Simplify

Multiply

8 - 4 = ? - 16

Calculate root

8 - 4 = ? - 4

Subtract term

4 \neq = - 4

The second solution is extraneous.

b Let's first rewrite the equations as functions by collecting all terms on one side of the equation and setting that side equal to

f (x)

and

g (x)

respectively.

x - 4 = 2 x : x - 4 = - 2 x : \Rightarrow f (x) = x - 4 - 2 x \Rightarrow g (x) = x - 4 + 2 x

Now we can graph the equation. Where they intersect the

x -

axis is a solution to the original equation.

As we can see from the diagram, the first equation has one solution at $x = 8$ and the other has a solution at $x = 2 .$

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268