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Mathematics 21 Online
OpenStudy (anonymous):

Solution to figure out a quadratic equation: 3, -2. Show work.

OpenStudy (anonymous):

if the solutions are 3 and -2, it means that the quadratic was like this: (x-3)(x+2)=0 expand that.

OpenStudy (anonymous):

If those are the roots then the factors are: (x-3)(x+2) Expand the brakcets \[y=x ^{2}-x-6\]

OpenStudy (anonymous):

cocnutshake, your equation is right, but y HAS to be equal to 0.

OpenStudy (anonymous):

So once you have y=x^2-x-6, how do you get Y to =0

OpenStudy (anonymous):

no no, use my equation.

OpenStudy (anonymous):

that's what i meant oops..

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

if the solution to the quadratic is 3, -2 it means that the quadratic was (x-3)(x+2)=0, only then, if you substitute x = 3 or x =-2 you will get LHS = RHS, which means only for this particular quadratic, 3,-2 are the solutions.

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

np

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