Question

If both x-2 and x - ½ are factors of px²+5x+q then find the relation of p and q.

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Answer 1

Given :-

(x - 2) and (x - ½) are the factors of px² + 5x + q

Let f(x) = px² + 5x + q

i) (x - 2) is a factor of px² + 5x + q

First find zero of (x - 2)

To find the zero of (x - 2) equate it to 0

x - 2 = 0

x = 2

By factor theorem f(2) = 0

$\sf f(2) = 0 \\ \\ \\ \sf \rightarrow p(2)^2 + 5(2) + q = 0 \\ \\ \\ \sf \rightarrow p(4) + 10 + q = 0 \\ \\ \\ \sf \rightarrow 4p + 10 + q = 0 \\ \\ \\ \sf \rightarrow 4p + q = - 10....(1)$

ii) (x - ½) is a factor of px² + 5x + q

First find zero of (x - ½)

To find the zero of (x - ½) equate it to 0

x - ½ = 0

x = ½

By factor theorem f(½) = 0

$\sf f \left( \dfrac{1}{2} \right) = 0 \\ \\ \\ \sf \rightarrow p \left( \dfrac{1}{2} \right)^2 + 5 \left( \dfrac{1}{2} \right) + q = 0 \\ \\ \\ \sf \rightarrow p \left( \dfrac{1}{4} \right) + \dfrac{5}{2} + q = 0 \\ \\ \\ \sf \rightarrow \dfrac{p}{4} + \dfrac{5}{2} + q = 0$

Taking LCM

$\sf \rightarrow \dfrac{p}{4} + \dfrac{5(2)}{2(2)} + \dfrac{q(4)}{1(4)} = 0 \\ \\ \\ \sf \rightarrow \dfrac{p}{4} + \dfrac{10}{4} + \dfrac{4q}{4} = 0 \\ \\ \\ \sf \rightarrow \dfrac{p + 10 + 4q}{4} = 0 \\ \\ \\ \sf \rightarrow p + 4q + 10 = 0 \times 4 \\ \\ \\ \sf \rightarrow p + 4q + 10 = 0 \\ \\ \\ \sf \rightarrow p + 4q = - 10......(2)$

From (1) and (2)

$\sf \rightarrow 4p + q = p + 4q \\ \\ \\ \sf \rightarrow 4p - p = 4q - q \\ \\ \\ \sf \rightarrow 3p = 3q \\ \\ \\ \sf \rightarrow p = q$

$\Huge{\boxed{ \tt p = q}}$