+0

# Algebra Question

0
424
3

Solve Equation. Check for extraneous solutions.

_____

a = 1 + (√7a -7)

Jun 12, 2017
edited by Guest  Jun 12, 2017

#1
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

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

Jun 12, 2017
#3
+7373
0

Definitely, you did something wrong.

MaxWong  Jun 13, 2017
#2
+7373
0

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

.
Jun 13, 2017