SOLUTION: {{{ -2x^2+50=0 }}}

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: {{{ -2x^2+50=0 }}}      Log On


   

Question 214129: +-2x%5E2%2B50=0+
Found 2 solutions by jim_thompson5910, drj:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

-2x%5E2%2B50=0 Start with the given equation.


Notice we have a quadratic in the form of ax%5E2%2Bbx%2Bc where a=-2, b=0, and c=50


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+=+%280+%2B-+sqrt%28+%280%29%5E2-4%28-2%29%2850%29+%29%29%2F%282%28-2%29%29 Plug in a=-2, b=0, and c=50


x+=+%280+%2B-+sqrt%28+0-4%28-2%29%2850%29+%29%29%2F%282%28-2%29%29 Square 0 to get 0.


x+=+%280+%2B-+sqrt%28+0--400+%29%29%2F%282%28-2%29%29 Multiply 4%28-2%29%2850%29 to get -400


x+=+%280+%2B-+sqrt%28+0%2B400+%29%29%2F%282%28-2%29%29 Rewrite sqrt%280--400%29 as sqrt%280%2B400%29


x+=+%280+%2B-+sqrt%28+400+%29%29%2F%282%28-2%29%29 Add 0 to 400 to get 400


x+=+%280+%2B-+sqrt%28+400+%29%29%2F%28-4%29 Multiply 2 and -2 to get -4.


x+=+%280+%2B-+20%29%2F%28-4%29 Take the square root of 400 to get 20.


x+=+%280+%2B+20%29%2F%28-4%29 or x+=+%280+-+20%29%2F%28-4%29 Break up the expression.


x+=+%2820%29%2F%28-4%29 or x+=++%28-20%29%2F%28-4%29 Combine like terms.


x+=+-5 or x+=+5 Simplify.


So the answers are x+=+-5 or x+=+5

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
+-2x%5E2%2B50=0+

Step 1. Add 2x%5E2 to both sides of equation to isolate the x-terms on side and the numbers on the other side.

+-2x%5E2%2B50%2B2x%5E2=0%2B2x%5E2+

50=2x%5E2

Step 2. Divide 2 to both sides of equation


50%2F2=2x%5E2%2F2


25=x%5E2

Step 3. Take square root of both sides of equation


sqrt%2825%29=sqrt%28x%5E2%29

5=x and -5=x

Step 4. The solutions to this equation are +5 and -5.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J