SOLUTION: Solve by graphing. If the solutions are not exact, estimate the answers by indicatinng the consecutive integer between which the root are located. x2-4x+3=0 x2+6x-2=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve by graphing. If the solutions are not exact, estimate the answers by indicatinng the consecutive integer between which the root are located. x2-4x+3=0 x2+6x-2=0       Log On


   

Question 23202: Solve by graphing. If the solutions are not exact, estimate the answers by indicatinng the consecutive integer between which the root are located.
x2-4x+3=0


x2+6x-2=0


-x2+6x-5-0


x2+ 7x-8= 0
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Do you have a graphing calculator? If so, just graph away:
x%5E2+-+4x+%2B3+=+0
graph%28300%2C300%2C-2%2C6%2C-10%2C10%2C+x%5E2-4x%2B3%29
As you can see, the roots are at x=1 and x=3

x%5E2+%2B6x+-2=0
graph%28300%2C300%2C-10%2C6%2C-10%2C10%2C+x%5E2%2B6x-2%29
As you can see, the roots are betweem x=-7 and x=-6, and between x=0 and x=1


-x%5E2+%2B6x+-5=0
graph%28300%2C300%2C-10%2C6%2C-10%2C10%2C+-x%5E2%2B6x-5%29
As you can see, the roots are at x-1 and x=5


x%5E2+%2B7x+-8=0
graph%28300%2C300%2C-10%2C6%2C-10%2C10%2C+x%5E2%2B7x-8%29
As you can see, the roots are at x=-8 and x=1


R^2 at SCC