SOLUTION: Prove that if {{{3x^2+6=18}}} and x<0, then x=-2 is true.

Algebra ->  Geometry-proofs -> SOLUTION: Prove that if {{{3x^2+6=18}}} and x<0, then x=-2 is true.       Log On


   

Question 188693: Prove that if 3x%5E2%2B6=18 and x<0, then x=-2 is true.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

3x%5E2%2B6=18 Start with the given equation.


3x%5E2%2B6-18=0 Subtract 18 from both sides.


3x%5E2-12=0 Combine like terms.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=3, b=0, and c=-12


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%280%29+%2B-+sqrt%28+%280%29%5E2-4%283%29%28-12%29+%29%29%2F%282%283%29%29 Plug in a=3, b=0, and c=-12


x+=+%280+%2B-+sqrt%28+0-4%283%29%28-12%29+%29%29%2F%282%283%29%29 Square 0 to get 0.


x+=+%280+%2B-+sqrt%28+0--144+%29%29%2F%282%283%29%29 Multiply 4%283%29%28-12%29 to get -144


x+=+%280+%2B-+sqrt%28+0%2B144+%29%29%2F%282%283%29%29 Rewrite sqrt%280--144%29 as sqrt%280%2B144%29


x+=+%280+%2B-+sqrt%28+144+%29%29%2F%282%283%29%29 Add 0 to 144 to get 144


x+=+%280+%2B-+sqrt%28+144+%29%29%2F%286%29 Multiply 2 and 3 to get 6.


x+=+%280+%2B-+12%29%2F%286%29 Take the square root of 144 to get 12.


x+=+%280+%2B+12%29%2F%286%29 or x+=+%28-0+-+12%29%2F%286%29 Break up the expression.


x+=+%2812%29%2F%286%29 or x+=++%28-12%29%2F%286%29 Combine like terms.


x+=+2 or x+=+-2 Simplify.


So the answers are x+=+2 or x+=+-2



However, since x%3C0, this means that "x" is negative. So this means that x=-2 is the only solution (if the inequality restriction is applied).