Question

If both (x-2) and (x-1/2)are the factors of px^2+5x+r,show that p=r

Answer 1

$\textbf{ Hello Mate! }$

As given, factors are : ( x - 2 ) and ( x - 1 / 2 ). So value of x = 2 and 1 / 2

p(x) = $p{x}^{2} + 5x + r$
0 = p(2)^2 + 5(2) + r
0 = 4p + 10 + r ______(1)

p(x) = p(1/2)^2 + 5(1/2) + r
0 = p / 4 + 5 / 2 + r
0 = ( p + 10 + 4r ) / 4
0 = p + 10 + 4r _____(2)

4p + 10 + r = p + 10 + 4r

4p + r = p + 4r

4p - p = 4r - r

3p = 3r

p = r

$\boxed{ Hence\:Proved }$

Have great future ahead!

Answer 2

Answer:

Hello Mate!

As given, factors are : ( x - 2 ) and ( x - 1 / 2 ). So value of x = 2 and 1 / 2

p(x) = p{x}^{2} + 5x + rpx

2

+5x+r

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

0 = 4p + 10 + r ______(1)

p(x) = p(1/2)^2 + 5(1/2) + r

0 = p / 4 + 5 / 2 + r

0 = ( p + 10 + 4r ) / 4

0 = p + 10 + 4r _____(2)

4p + 10 + r = p + 10 + 4r

4p + r = p + 4r

4p - p = 4r - r

3p = 3r

p = r

\boxed{ Hence\:Proved }

HenceProved