SOLUTION: what are the cordinates of this equation x^2-x-12=0

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: what are the cordinates of this equation x^2-x-12=0      Log On


   

Question 603572: what are the cordinates of this equation x^2-x-12=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2-x-12=0 Start with the given equation.


Notice that the quadratic x%5E2-x-12 is in the form of Ax%5E2%2BBx%2BC where A=1, B=-1, 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-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2-4%281%29%28-12%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-1, and C=-12


x+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%281%29%28-12%29+%29%29%2F%282%281%29%29 Negate -1 to get 1.


x+=+%281+%2B-+sqrt%28+1-4%281%29%28-12%29+%29%29%2F%282%281%29%29 Square -1 to get 1.


x+=+%281+%2B-+sqrt%28+1--48+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-12%29 to get -48


x+=+%281+%2B-+sqrt%28+1%2B48+%29%29%2F%282%281%29%29 Rewrite sqrt%281--48%29 as sqrt%281%2B48%29


x+=+%281+%2B-+sqrt%28+49+%29%29%2F%282%281%29%29 Add 1 to 48 to get 49


x+=+%281+%2B-+sqrt%28+49+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%281+%2B-+7%29%2F%282%29 Take the square root of 49 to get 7.


x+=+%281+%2B+7%29%2F%282%29 or x+=+%281+-+7%29%2F%282%29 Break up the expression.


x+=+%288%29%2F%282%29 or x+=++%28-6%29%2F%282%29 Combine like terms.


x+=+4 or x+=+-3 Simplify.


So the solutions are x+=+4 or x+=+-3