Question

if both (x-2) and (x-1/2) are the factors of px2+5x+r then show that p/r=1​

Answer 1

Step-by-step explanation:

Here.

$f(x)= p {x}^{2} + 5x + r$

Now. x-2 is a factor the f(x) will zero at x= 2

$f(2) = 0 \\ \\ p(2) {}^{2} + 5(2) + r = 0 \\ \\ 4p + 10 + r = 0 \\ \\ 4p + r = - 10 \: \: \: ... ...... .....(1)$

Similarly for x-1/2

$f( \frac{1}{2} ) = 0 \\ \\ p( \frac{1}{2} ) {}^{2} + 5( \frac{1}{2} ) + r = 0 \\ \\ \frac{p}{4} + \frac{ 5}{2} + r = 0 \\ \\ \frac{p}{4} + r = - \frac{5}{2} \: \: \: \: \: ..............(2)$

subtract 1 and 2

$4p - \frac{p}{4} = - 10 + \frac{5}{2} \\ \\ \frac{15p}{4} = \frac{ - 15}{2} \\ \\ p = - 2$

Now put p= -2 in eq 1

$( - 2) \times 4 + r = - 10\\ - 8 + r = - 10 \\ r = - 10 + 8 \\ r = - 2$

now

$\frac{p }{r} = \frac{ - 2}{ - 2} = 1$

Answer 2

As both are the factors of the given polynomials, both must be zero for x = 2 and 1/2.   Using factor theorem:

 If x - 2 is factor: f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + 10 + r = 0          ...(1)

 If x - 1/2 is factor: f(1/2) = 0

⇒ p(1/2)² + 5(1/2) + r = 0

⇒ p/4 + 5/2 + r = 0

⇒ p + 10 + 4r = 0         ...(2)

   Subtract (1) from (2), we get:

⇒ 3p - 3r = 0

⇒ 3p = 3r

⇒ p = r      proved