Question

Pls 8th question need fast

Answer 1

Hope u like my process
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=> f(x) = px² + 5x + r

=> If (x-2) and (x - ½) are the factors of f(x),
then,

f(x) =0, when x = 2, ½

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By problem
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$= > p {x}^{2} + 5x + r = 0.....(when \: \: x = 2) \\ \\ = > p {(2)}^{2} + 5(2) + r = 0 \\ \\ = > 4p + r = - 10.......(1) \\ \\ \bf \: \underline{now} \\ \\ = > p {x}^{2} + 5x + r = 0......(when \: \: x = \frac{1}{2} ) \\ \\ = > p {( \frac{1}{2}) }^{2} + 5 ( \frac{1}{2} ) + r = 0 \\ \\ = > 4( \frac{p}{4 } + \frac{5}{2} + r) = 4 \times 0 \\ \\ = > p + 10 + 4r = 0 \\ \\ = > p + 4r = - 10........(2)$
Now comparing equation (1) & (2)
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$= > 4p + r = p + 4r \\ \\ = > 4p - p = 4r - r \\ \\ = > 3p = 3r \\ \\ = > \bf \: p = r \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \underline{(proved)}$
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Hope this is ur required answer

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