+0  
 
0
38
3
avatar

Solve Equation. Check for extraneous solutions.

             _____

a = 1 + (√7a -7)

Guest Jun 12, 2017
edited by Guest  Jun 12, 2017
Sort: 

3+0 Answers

 #1
avatar
0

I have taken 7a under the square root sign. I also don't know what you mean by "extraneous solutions!"

 

Solve for a:
a = sqrt(7) sqrt(a) - 6

Subtract sqrt(7) sqrt(a) + a from both sides:
-sqrt(7) sqrt(a) = -a - 6

Raise both sides to the power of two:
7 a = (-a - 6)^2

Expand out terms of the right hand side:
7 a = a^2 + 12 a + 36

Subtract a^2 + 12 a + 36 from both sides:
-a^2 - 5 a - 36 = 0

Multiply both sides by -1:
a^2 + 5 a + 36 = 0

Subtract 36 from both sides:
a^2 + 5 a = -36

Add 25/4 to both sides:
a^2 + 5 a + 25/4 = -119/4

Write the left hand side as a square:
(a + 5/2)^2 = -119/4

Take the square root of both sides:
a + 5/2 = (i sqrt(119))/2 or a + 5/2 = -(i sqrt(119))/2

Subtract 5/2 from both sides:
a = (i sqrt(119))/2 - 5/2 or a + 5/2 = -(i sqrt(119))/2

Subtract 5/2 from both sides:
Answer: | a = (i sqrt(119))/2 - 5/2      or        a = -(i sqrt(119))/2 - 5/2

Guest Jun 12, 2017
 #3
avatar+6646 
0

Definitely, you did something wrong.

MaxWong  Jun 13, 2017
 #2
avatar+6646 
0

\(a=1+\sqrt{7a-7}\\ a-1=\sqrt{7a-7}\\ 7a-7=(a-1)^2\\ a^2 - 2a + 1 = 7a - 7\\ a^2 - 9a + 8 = 0\\ (a-1)(a-8)=0\\ a=1\text{ or }8\)

MaxWong  Jun 13, 2017

6 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details