SOLUTION: My textbook for Algebra asks for us to "Solve the equation" but our teacher didn't go over this and I'm confused. The problem in question is {{{ 1/x^2 + 8/x +15=0}}} am I supposed

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: My textbook for Algebra asks for us to "Solve the equation" but our teacher didn't go over this and I'm confused. The problem in question is {{{ 1/x^2 + 8/x +15=0}}} am I supposed       Log On


   



Question 845307: My textbook for Algebra asks for us to "Solve the equation" but our teacher didn't go over this and I'm confused. The problem in question is +1%2Fx%5E2+%2B+8%2Fx+%2B15=0 am I supposed to use the quadratic formula for this? Thanks for your trouble!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You could go about it one of two ways:
1) You could multiply both sides of the equation times x%5E2 to get
1%2B8x%2B15x%5E2=0 <--> 15x%5E2%2B8x%2B1=0 and solve that equation.
(That is probably the expected way).
2) You could say y=1%2Fx , and that makes y%5E2=1%2Fx%5E2 ,
and substituting you would get
y%5E2%2B8y%2B15=0 .
Solving that equation you get solutions for y ,
which you can substitute into y=1%2Fx to find the values for x.

MULTIPLYING TIMES x%5E2:
We know that x%5E2 is not zero, because it is in a denominator and if it were zero, 1%2Fx%5E2 would not exist. If we found x=0 as a solution, we would discard it, because we know it is not a valid solution of
+1%2Fx%5E2+%2B+8%2Fx+%2B15=0 .
So we can safely multiply times x%5E2 both sides of the equal sign.will get
+1%2Fx%5E2+%2B+8%2Fx+%2B15=0
x%5E2%2A%281%2Fx%5E2+%2B+8%2Fx+%2B15%29=x%5E2%2A0
+x%5E2%281%2Fx%5E2%29+%2B+x%5E2%2A%288%2Fx%29+%2Bx%5E2%2A15=0
1%2B8x%2B15x%5E2=0
15x%5E2%2B8x%2B1=0
The quadratic formula says that the solutions to
ax%5E2%2Bbx%2Bc=0 are given by
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
15x%5E2%2B8x%2B1=0 is ax%5E2%2Bbx%2Bc=0 with
a=15 , b=8 , and c=1 .
So x+=+%28-8+%2B-+sqrt%288%5E2-4%2A15%2A1+%29%29%2F%282%2A15%29+
x+=+%28-8+%2B-+sqrt%2864-60%29%29%2F30+
x+=+%28-8+%2B-+sqrt%284%29%29%2F30+
x+=+%28-8+%2B-+2%29%2F30+
The solutions are
x+=+%28-8+%2B+2%29%2F30+=-6%2F30=highlight%28-1%2F5%29 and x+=+%28-8+-+2%29%2F30+=-10%2F30=highlight%28-1%2F3%29

NOTE:
You could use the quadratic formula, or you could solve by factoring, or you could even solve by completing the square.
For 15x%5E2%2B8x%2B1=0 , using the quadratic formula is very reasonable,
because the other options do not seem any easier.
If you had changed variables (option 2) above) to end with
y%5E2%2B8y%2B15=0 ,
you may find that factoring is the easiest option.
You may immediately see that you can factor to get
%28y%2B3%29%28y%2B5%29=0 ,
which would tell you that the solutions are y=-3 and y=-5 .
y=-3 means 1%2Fx=-3 --> 1=-3x -->1%2F%28-3%29=x --> x=-1%2F3
y=-5 means 1%2Fx=-5 --> 1=-5x -->1%2F%28-5%29=x --> x=-1%2F5