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

$$\mathrm{Let}\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{be}\mathrm{two}\mathrm{positive}\mathrm{real}\mathrm{number}$$$$\mathrm{such}\mathrm{that}\{\begin{array}{l}\mathrm{p}\sqrt{\mathrm{p}}+\mathrm{q}\sqrt{\mathrm{q}}=32\\ \mathrm{p}\sqrt{\mathrm{q}}+\mathrm{q}\sqrt{\mathrm{p}}=31\end{array}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{5(\mathrm{p}+\mathrm{q})?}{7}$$

Answered by mindispower last updated on 01/Jan/21

$${(\sqrt{p}+\sqrt{q})}^3=p\sqrt{p}+q\sqrt{q}+3(p\sqrt{q}+p\sqrt{q})=32+3.31=125$$$$\Rightarrow \sqrt{p}+\sqrt{q}=5$$$$2ndEq\iff \sqrt{pq}.(\sqrt{p}+\sqrt{q})=31$$$$\Rightarrow \sqrt{pq}=\frac{31}{5}$$$$p+q={(\sqrt{p}+\sqrt{q})}^2-2\sqrt{pq}$$$$=5^2-2.\frac{31}{5}=\frac{125-62}{5}=\frac{63}{5}$$$$\frac{5}{7}(p+q)=9$$$$$$

Answered by bemath last updated on 01/Jan/21

$$\{\begin{array}{l}\mathrm{p}\sqrt{\mathrm{p}}+\mathrm{q}\sqrt{\mathrm{q}}=32...\left(\mathrm{i}\right)\\ \mathrm{p}\sqrt{\mathrm{q}}+\mathrm{q}\sqrt{\mathrm{p}}=31...\left(\mathrm{ii}\right)\end{array}$$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow \mathrm{p}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})+\mathrm{q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})=63$$$$\Rightarrow (\mathrm{p}+\mathrm{q})(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})=63...\left(\mathrm{iii}\right)$$$$\mathrm{squaring}\Rightarrow {(\mathrm{p}+\mathrm{q})}^2{(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})}^2={63}^2...\left(\mathrm{iv}\right)$$$$\mathrm{consider}\mathrm{eq}\left(\mathrm{ii}\right):\sqrt{\mathrm{pq}}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})=31$$$$\mathrm{we}\mathrm{get}\frac{63}{\mathrm{p}+\mathrm{q}}=\frac{31}{\sqrt{\mathrm{pq}}}\mathrm{or}\sqrt{\mathrm{pq}}=\frac{31}{63}(\mathrm{p}+\mathrm{q})...\left(\mathrm{v}\right)$$$$\mathrm{we}\mathrm{know}\mathrm{that}{(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})}^2=\mathrm{p}+\mathrm{q}+2\sqrt{\mathrm{pq}}$$$${(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})}^2=\mathrm{p}+\mathrm{q}+\frac{62}{63}(\mathrm{p}+\mathrm{q})$$$${(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})}^2=\frac{125}{63}(\mathrm{p}+\mathrm{q})...\left(\mathrm{vi}\right)$$$$\mathrm{substitute}\mathrm{into}\mathrm{eq}\left(\mathrm{iv}\right)\mathrm{gives}$$$${(\mathrm{p}+\mathrm{q})}^2\times \frac{125}{63}(\mathrm{p}+\mathrm{q})={63}^2$$$${(\mathrm{p}+\mathrm{q})}^3=\frac{{63}^3}{5^3}\Rightarrow \mathrm{p}+\mathrm{q}=\frac{63}{5}$$$$\mathrm{therefore}\frac{5(\mathrm{p}+\mathrm{q})}{7}=\frac{5}{7}\times \frac{63}{5}=9$$$$$$

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