Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

811,112 users

SOMEONE MAJOR HELP!!!!! ANALYSIS & FUNCTIONS! 10 points for best answer!?

0 votes
7.
Solve this equation
squrt x+3 = x - 3

8.
solve this equation
squrt 6a -8 - sqrt 2a-9 = 0
asked Jun 7, 2013 in ALGEBRA 2 by linda Scholar

2 Answers

0 votes

7.The equation is squrt x+3 = x - 3.

Apply squre  each side.

(√x+3)^2 = (x-3)^2

x+3 = x^2-2(x)(3)+3^2

x+3 = x^2-6x+9

Subtract (x+3) from each side.

x+3 - (x+3) = x^2-6x+9-(x+3)

0 = x^2-6x+9-x-3

x^2-7x+6 = 0

x^2-6x-x+6 = 0

x(x-6)-1(x-6) = 0

Take out common factors.

(x-1)(x-6) = 0

Apply zero product property

x - 1= 0 or x - 6 = 0

x = 1 or x = 6

The solution is x = 1 or 6.

answered Jun 7, 2013 by kevin Rookie

The radical equation is √(x + 3) = x - 3.

The solutions are x = 1 and x = 6.

It is very important that you check your solution. Sometimes you will obtain a number that does not satisfy the original equation. Such a number is called an extraneous solution.

To check the solution, substitute the value of x = 1 in the original equation √(x + 3) = x - 3.

√[ (1) + 3 ] ≟ (1) - 3

√(4) ≟ - 2

2 = - 2

The above statement is false, so the value of x = 1 is not a solution of the original equation.

To check the solution, substitute the value of x = 6 in the original equation √(x + 3) = x - 3.

√[ (6) + 3 ] ≟ (6) - 3

√(9) ≟ 3

3 = 3

The above statement is true, so the value of x = 6 is a solution of the original equation.

Therefore, solution of the equation √(x + 3) = x - 3 is 6.

0 votes

8.The is equation is √( 6a -8) - √( 2a-9) = 0

Add √(2a-9) to each side

√( 6a-8) -√( 2a-9) + √( 2a-9) =√( 2a-9)

√(6a-8 )= √(2a-9)

apply square each side

6a-8 = 2a-9

subtract (2a  - 9) from each side

6a-8-(2a-9) = 2a-9-(2a-9)

6a-8-2a+9 = 2a-9-2a+9

4a+1 = 0

4a = -1

a =-1/4

The solution is a =-1/4.

answered Jun 7, 2013 by kevin Rookie

Related questions

...