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Solve Equation. Check for extraneous solutions.

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a = 1 + (√7a -7)

Guest 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

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