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

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) (Show Source): You can put this solution on YOUR website!
The value for P does not matter - Maybe.

The question just specified REAL values of P, Q, and R.
.

The entire system can be multiplied by any real number factor and will still work.
Your equation can also be .
Answer by MathTherapy(10552) (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)

With , it's lucid that:
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