Question

A is equals to root 3 minus root 2 upon root 3 + root 2 b is equals to root 3 + root 2 upon root 3 minus root 2 find a square + b square minus 5 a b

Answer 1

Concept Introduction:-

It may be in the form of a word, a symbol, or a figure that reflects the arithmetic value of a quantity.

Given Information:-

We have been given that $a$ is equals to root $3$ minus root $2$ upon root $3$ $+$ root $2$ $b$ is equals to root $3$ $+$ root $2$ upon root $3$ minus root $2$

To Find:-

We have to find that the value of $a$ square $+ b$ square minus $5 a b$

Solution:-

According to the problem

$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$a b=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}$

$a b=1$

now

$a=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\\a=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \frac{(\sqrt{3}-\sqrt{3})}{(\sqrt{2})}\\a=(\sqrt{3}-\sqrt{2})^{2}\\a=5-2 \sqrt{6}$

$b=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\\b=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})} \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\\b=(\sqrt{3}+\sqrt{2})^{2}\\a=5-2 \sqrt{6}\\b=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\\b=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})} \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\\b=(\sqrt{3}+\sqrt{2})^{2}\\b=5+2 \sqrt{6}\\a^{2}=49-20 \sqrt{6}\\b^{2}=49+20 \sqrt{6}$

now

$a^{2}+b^{2}-5 a b\\=93$

Final Answer:-

The value of $a$ square $+ b$ square minus $5 a b$ is $93$.

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