Question 63348
solve. 
:
a)x^2 - 2x = 15
: 
Subtract 15 to both sides, we have simple quadratic equation, we can factor:
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = +5
and
x = -3
:
Substituting either solution in the original equation will prove our solutions
:
:
b)3x^2 + 19 + 1 = 0
3x^2 + 20 = 0
:
Subtract 20 from both sides:
3x^2 = -20
:
Divide both sides by 3
x^2 = -20/3
x = +/- SqRt(-20/3)
x =+/- SqRt((5*4*-1)/3)
:
Extract the SqRt(4) and i (sqRt of -1):
x = +/- 2i*SqRt(5/3)
:
:
:
c){{{1/(4x+12) - 1/(x^2-9) =  5/(x-3)}}}
:
Factor where we can:
{{{1/(4(x+3)) - 1/((x-3)(x+3)) =  5/(x-3)}}}
:
We can see the common denominator is 4(x+3)(x-3), mult eq by that and we get:
1(x-3) - 4(1) = 4(x+3)*5
x - 3 - 4 = 20x + 60
x - 7 = 20x + 60
x - 20x = 60 + 7
-19x = + 67
x = -67/19