# How do you solve these radical equations?

+1 vote

1. simplify the expression. 5-√6 ÷ 8-√6

2. simplify the expression. make all exponents positive. (16x^-12(y^24))^1/2

3. solve. check for extraneous solutions. (2x^4/3)-5=27

4. solve. check for extraneous solutions. √(2x+7)=x+2

All help will be appreciated :)

+1 vote

1). 5-√6 ÷ 8-√6 = ?

To simplify the expression follow the order of operations: (PEMDAS)

1. Evaluate the expressions inside Brackets/Parenthesis (P).
2. Evaluate Powers/Exponents (E).
3. Multiply and/or Divide in order from left to right (MD).
4. Add and/or Subtract in order from left to right (AS).

A simple technique for remembering the order PE operations is turned into phrase

“Please Excuse My Dear Aunty Sally”

First operation is Evaluate the expressions Parenthesis.

According to division property of equality:

5-√6 ÷ 8-√6 = 5 - √6/8 - √6

The least common denominator (LCD) is 8.

= [(5)(8) - (√6) - (8)(√6)] / 8

= (40 - √6 - 8√6) / 8

= (40 - 9√6) / 8

+1 vote

(16x^-12)(y^24))^1/2

= [ (16x-12)(y24)1/2 ]

Recall identities and properties is

bm+n = bm·bn, (bm)n = bm.n, (an)1/n= a and  (b·c)n = bn·cn

= [ (4)2(x-6)2(y12)2]1/2

= { [ (4x-6)(y12)]2}1/2

Simplify

= (4x-6)(y12)

+1 vote

2x4/3 - 5 = 27

2x4/3 - 5 + 5 = 27 + 5

Simplify

2x4/3 = 32

Divide each side by 2.

( 2x4/3) / 2  = 32 / 2

Simplify

x4/3 = 16

(x1/3)4 = 24

x1/3 = 2

Apply cube each side.

(x1/3)3 = 23

x = 8.

Check for extraneous solutions is (2x^4/3)-5=27

2(8)4/3 - 5 = 2(23)4/3 - 5 = 2(2)4 - 5 = 2(16) - 5 = 32 - 5 = 27

+1 vote

√(2x+7) = x+2

Apply square each side.

[√(2x+7)]2 = [x + 2]2

Recall formula (√a)2 = a and (a + b)2 = a2+2ab+b2

So, 2x + 7 = x2 + 2(x)(2) + 22

2x + 7 = x2 + 4x + 4

Subtract 2x from each side.

2x + 7 - 2x = x2 + 4x + 4 - 2x

7 = x2 + 2x + 4

Subtract 7 from each side.

7 - 7 = x2 + 2x + 4 - 7

x2 + 2x - 3 = 0

Now solve the factor method.

x2 + 3x - x - 3 = 0

x(x + 3)- 1(x + 3) = 0

Take out common factors.

(x + 3)(x - 1) = 0

x + 3 = 0

Subtract 3 from each side.

x = -3

and x - 1 = 0

x = 1

There fore x = 1 or - 3

If x =1

check for extraneous solutions is √(2x+7)=x+2

√(2(1)+7)=1+2

√(2+7)=3

√(9)= 3

3 = 3

If x = -3

check for extraneous solutions is √(2x+7)=x+2

√(2(-3)+7)=-3+2

√(-6+7)= -1

√(1)= -1

+ 1 = -1