Question

if both x-2 and x-1/2 are factors of px2+5x+r show that p=r

Answer 1

   

f(x) = px^2 + 5x + r = 0

(x - 2) and (x - 1/2) are the factors. So 2 and 1/2 are the roots.

f(2) = p(2^2) + 5(2) + r = 0

f(2) = p(4) + 5 × 2 + r = 0

f(2) = 4p + 10 + r = 0 --> (1)

f(1/2) = p((1/2)^2) + 5(1/2) + r = 0

f(1/2) = p(1/4) + 5 × 1/2 + r = 0

f(1/2) = p/4 + 5/2 + r = 0 --> (2)

Here, (1) = (2)

=> 4p + 10 + r = p/4 + 5/2 + r

=> 4p + 10 = p/4 + 5/2

=> 4(4p + 10) = 4(p/4 + 5/2)

=> 16p + 40 = p + 10

=> 16p - p = 10 - 40

=> 15p = -30

=> p = -2

From (1),

=> 4p + 10 + r = 0

=> 4(-2) + 10 + r = 0

=> -8 + 10 + r = 0

=> 2 + r = 0

=> r = -2

So,  p = r = -2.

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