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Question Number 202715 by Rasheed.Sindhi last updated on 01/Jan/24

$$\overline{)\begin{array}{c}\underset{\mathrm{is}2025}{\mathrm{Square}\mathrm{of}\mathrm{mean}\mathrm{of}\mathrm{two}\mathrm{numbers}}\end{array}}$$$$\mathrm{\&}$$$$\overline{)\begin{array}{c}\underset{\mathrm{is}2026}{\mathrm{mean}\mathrm{of}\mathrm{squares}\mathrm{of}\mathrm{the}\mathrm{numbers}}\end{array}}$$$$$$$$\overline{)\begin{array}{c}\mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{numbers}.\end{array}}$$

Answered by JDamian last updated on 01/Jan/24

$${\left(\frac{x+y}{2}\right)}^2=2025=\frac{1}{2}(\frac{x^2+y^2}{2}+xy)$$$$\frac{x^2+y^2}{2}=2026$$$$4050=2026+xy$$$$xy=4050-2026=2024$$

Commented by Rasheed.Sindhi last updated on 01/Jan/24

$$\mathcal{T}hankssir!$$$$2024✓$$$$\overline{)\begin{array}{c}\stackrel{\mathbb{HAPPY}}{\mathbb{NEW}\mathbb{YEAR}!}\end{array}}$$

Answered by professorleiciano last updated on 01/Jan/24

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