SOLUTION: find the roots of each equation by completing the square. show your work. 1. x^2-6x-10=0 2. 3x^2-6x-6

Algebra ->  Rational-functions -> SOLUTION: find the roots of each equation by completing the square. show your work. 1. x^2-6x-10=0 2. 3x^2-6x-6      Log On


   



Question 926519: find the roots of each equation by completing the square. show your work.
1. x^2-6x-10=0
2. 3x^2-6x-6

Found 2 solutions by EMStelley, MathLover1:
Answer by EMStelley(208) About Me  (Show Source):
You can put this solution on YOUR website!
The steps to completing the square are the same for each of these, but the first one contains one less step since the coefficient of x%5E2 is 1.
For #1, the first step is to move the constant to the other side.
x%5E2+-+6x+=+10
Now, we take the b term (-6 in this case), divide it by half and square it, and add it to both sides: (b/2)^2.
x%5E2+-+6x+%2B+%28-6%2F2%29%5E2+=+10+%2B+%28-6%2F2%29%5E2
Now, this past step created what is called a perfect square trinomial, which allows us to factor the left hand side as %28x+%2B+%28b%2F2%29%29%5E2, so we have:
%28x+%2B+%28-6%2F2%29%29%5E2+=+10+%2B+9
Simplifying a bit we get
%28x+-+3%29%5E2+=+19
Since we now have something squared equal to a constant we can take the square root of both sides to obtain:
x+-+3+=+%2B-+sqrt%2819%29
Adding 3 to both sides we get our final answer:
x+=+3+%2B-+sqrt%2819%29
That's it!!
The only difference in #2 is that we must divide through by 3 in order for the coefficient of the x%5E2 term to be 1. So, step by step we have:
3x%5E2+-+6x+=+6
x%5E2+-+2x+=+2
x%5E2+-+2x+%2B+%28-2%2F2%29%5E2+=+2+%2B+%28-2%2F2%29%5E2
%28x+%2B+%28-2%2F2%29%29%5E2+=+2+%2B+1
%28x+-+1%29%5E2+=+3
x+-+1+=+%2B-+sqrt%283%29
x+=+1+%2B-+sqrt%283%29
Hope that helps!!

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

1.
x%5E2-6x-10=0
%28x%5E2-6x%2B___%29-_____-10=0
%28x%5E2-6x%2B_3%5E2_%29-__3%5E2__-10=0
%28x-3%29%5E2-9-10=0
%28x-3%29%5E2-19=0
%28x-3%29%5E2=19
%28x-3%29=sqrt%2819%29.....sqrt%2819%29%5E2 is sqrt%2819%29 or -sqrt%2819%29
solutions:
%28x-3%29=sqrt%2819%29
x=3%2Bsqrt%2819%29
or
%28x-3%29=-sqrt%2819%29
x=3-sqrt%2819%29


2.
3x%5E2-6x-6=0
%283x%5E2-6x%29-6=0
3%28x%5E2-2x%29-6=0
3%28x%5E2-2x%2B1%5E2%29-3%2A1%5E2-6=0
3%28x-1%29%5E2-3-6=0
3%28x-1%29%5E2-9+=+0
3%28x-1%29%5E2+=+9
%28x-1%29%5E2+=+9%2F3
%28x-1%29%5E2+=+3
%28x-1%29+=+sqrt%283%29
solutions:
x-1+=+sqrt%283%29
x+=1%2B+sqrt%283%29
or
x-1+=+-sqrt%283%29
x+=1-+sqrt%283%29