Solve Equation. Check for extraneous solutions.
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a = 1 + (√7a -7)
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