If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then:

Question

If both (x – 2) and (x – \(\frac{1}{2}\)) are factors of px² + 5x + r, then:

(a) p = 2r 

(b) p + r = 0 

(c) p = r 

(d) p x r = 1

Answer 1

(c) p = r.

Let f(x) = px² + 5x + r 

Since, (x – 2) and  \(\big(\)x – \(\frac{1}{2}\)\(\big)\) are the factors of f(x), therefore, 

f(2) = 0 and f = \(\big(\frac{1}{2}\big)\) = 0. 

f(2) = p × (2)² + 5 × 2 + r = 4p + 10 + r = 0                    ...(i) 

f\(\big(\frac{1}{2}\big)\) = p x \(\big(\frac{1}{2}\big)\)² + 5 x \(\frac{1}{2}\) + r = \(\frac{p}{4} + \frac{5}{2}\) + r = p + 10 + 4r = 0      ...(ii) 

(i) and (ii) ⇒ 4p + 10 + r = p + 10 + 4r ⇒ 3p = 3r ⇒ p = r.