Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

  
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

807,706 users

math please!! 12

0 votes

solve a) √(7x)=5    b)√(4x-5)=6     c) √6x+7 = √ 7x+7+6

asked Nov 21, 2014 in PRECALCULUS by anonymous

3 Answers

0 votes

a) The radical equation √(7x) = 5

Squaring on each side.

[√(7x)]2 = 52

7x = 25

Divide each side by 7.

x = 25/7

Solution x = 25/7.

answered Nov 21, 2014 by david Expert
0 votes

b) The radical equation √(4x - 5) = 6

Squaring on each side.

[√(4x - 5)]2 = 62

4x - 5 = 36

4x = 36 + 5

4x = 41

x = 41/4

Solution x = 41/4.

answered Nov 21, 2014 by david Expert
0 votes

(c)

√(6x+7) = √ (7x+7) + 6

Square on both sides.

(√(6x+7))² = (√(7x+7) + 6)²

6x + 7 = 7x + 7 + 36 +12√(7x+7)

7x - 6x + 36 =  - 12√(7x+7)

x + 36 = -12√(7x+7)

Square on both sides.

(x + 36)² = 144(7x + 7)

x² + 1296 + 72x = 1008x + 1008

x² - 936x + 288 = 0

Roots of the Equation

image

image

image

image

image

Therefore the values of x are 468 ± 84√31.

answered Nov 21, 2014 by Lucy Mentor

Related questions

asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Nov 15, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Dec 19, 2014 in PRECALCULUS by anonymous
...