+0  
 
0
379
3
avatar

Solve Equation. Check for extraneous solutions.

             _____

a = 1 + (√7a -7)

Guest Jun 12, 2017
edited by Guest  Jun 12, 2017
 #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+7026 
0

Definitely, you did something wrong.

MaxWong  Jun 13, 2017
 #2
avatar+7026 
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

46 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.