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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 :)

asked Jan 29, 2013 in PRE-ALGEBRA by andrew Scholar

4 Answers

+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

answered Jan 29, 2013 by richardson Scholar
+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)

answered Jan 29, 2013 by richardson Scholar
+1 vote

2x4/3 - 5 = 27

Add 5 to each side.

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

answered Jan 29, 2013 by richardson Scholar
+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

Add 1 to each side

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

answered Jan 29, 2013 by richardson Scholar

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