SOLUTION: Given y=Px^2+Qx+R,find the values of P,A and R,when the roots x are -2 and 0 .5.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Given y=Px^2+Qx+R,find the values of P,A and R,when the roots x are -2 and 0 .5.      Log On


   

Question 954033: Given y=Px^2+Qx+R,find the values of P,A and R,when the roots x are -2 and 0 .5.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The value for P does not matter - Maybe.

%28x%2B2%29%28x-1%2F2%29=0
x%5E2%2B2x-x%2F2-1=0
x%5E2%2B%283%2F2%29x-1=0

The question just specified REAL values of P, Q, and R.
system%28P=1%2CQ=3%2F2%2CR=-1%29.

The entire system can be multiplied by any real number factor and will still work.
Your equation can also be y=2x%5E2%2B3x-2.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Given y=Px^2+Qx+R,find the values of P,A and R,when the roots x are -2 and 0 .5.
You must mean: P, Q, R since there's no "A" in the equation.



x = - 2

x + 2 = 0

Therefore, x + 2 is a factor of the trinomial



x = .5

x - .5 = 0 

2x – 1 = 0 ------- Multiplying by 2

Therefore, 2x - 1 is a factor of the trinomial



We now have: y = (x + 2)(2x – 1)

y+=+2x%5E2+%2B+3x+-+2

With y+=+Px%5E2+%2B+Qx+%2B+R, it's lucid that: highlight_green%28system%28P+=+2%2CQ+=+3_and%2CR+=+-+2%29%29