Let-p-and-q-be-two-positive-real-number-such-that-p-p-q-q-32-p-q-q-p-31-find-the-value-of-5-p-q-7-

Question Number 127716 by liberty last updated on 01/Jan/21

Let p and q be two positive real number

such that {\begin{cases} p \sqrt{p} + q \sqrt{q} = 32 \\ p \sqrt{q} + q \sqrt{p} = 31 \end{cases}

find the value of \frac{5 (p + q) ?}{7}

Answered by mindispower last updated on 01/Jan/21

{(\sqrt{p} + \sqrt{q})}^{3} = p \sqrt{p} + q \sqrt{q} + 3 (p \sqrt{q} + p \sqrt{q}) = 32 + 3.31 = 125

\Rightarrow \sqrt{p} + \sqrt{q} = 5

2 n d E q \Leftrightarrow \sqrt{p q} . (\sqrt{p} + \sqrt{q}) = 31

\Rightarrow \sqrt{p q} = \frac{31}{5}

p + q = {(\sqrt{p} + \sqrt{q})}^{2} - 2 \sqrt{p q}

= 5^{2} - 2 . \frac{31}{5} = \frac{125 - 62}{5} = \frac{63}{5}

\frac{5}{7} (p + q) = 9

Answered by bemath last updated on 01/Jan/21

{\begin{cases} p \sqrt{p} + q \sqrt{q} = 32 \dots (i) \\ p \sqrt{q} + q \sqrt{p} = 31 \dots (ii) \end{cases}

(i) + (ii) \Rightarrow p (\sqrt{p} + \sqrt{q}) + q (\sqrt{p} + \sqrt{q}) = 63

\Rightarrow (p + q) (\sqrt{p} + \sqrt{q}) = 63 \dots (iii)

squaring \Rightarrow {(p + q)}^{2} {(\sqrt{p} + \sqrt{q})}^{2} = 63^{2} \dots (iv)

consider eq (ii) : \sqrt{pq} (\sqrt{p} + \sqrt{q}) = 31

we get \frac{63}{p + q} = \frac{31}{\sqrt{pq}} or \sqrt{pq} = \frac{31}{63} (p + q) \dots (v)

we know that {(\sqrt{p} + \sqrt{q})}^{2} = p + q + 2 \sqrt{pq}

{(\sqrt{p} + \sqrt{q})}^{2} = p + q + \frac{62}{63} (p + q)

{(\sqrt{p} + \sqrt{q})}^{2} = \frac{125}{63} (p + q) \dots (vi)

substitute into eq (iv) gives

{(p + q)}^{2} \times \frac{125}{63} (p + q) = 63^{2}

{(p + q)}^{3} = \frac{63^{3}}{5^{3}} \Rightarrow p + q = \frac{63}{5}

therefore \frac{5 (p + q)}{7} = \frac{5}{7} \times \frac{63}{5} = 9

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