Question

Please help me in this question with steps

Answer 1

Answer:

Ans. = 14

Step-by-step explanation:

Here it is frnd !!

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Answer 2

✪ $\huge{\red{\underline{\overline{\mathbf{Question}}}}}$ ✪

→$if \:x=2+\sqrt{3} , \:find \: the\:value\:of\:x^2+\frac{1}{x^2}$

✪ $\huge{\green{\underline{\overline{\mathbf{Answer}}}}}$ ✪

⇒Given:

• $x=2+\sqrt{3}$

⇒To Find:

• $x^2+\frac{1}{x^2}$

⇒Solution:

⇒$x=2+\sqrt{3}$

On squaring both sides:-

⇒$x^2=(2+\sqrt{3})^2$

⇒$x^2=4+3+4\sqrt{3}$ [using formula]

⇒$x^2=7+4\sqrt{3}$

Now-

⇒$\frac{1}{x^2}=\frac{1}{7+4\sqrt{3}}$

⇒$\frac{1}{x^2}=\frac{1}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}$

⇒$\frac{1}{x^2}=\frac{7-4\sqrt{3}}{7^2-(4\sqrt{3})^2}$ [Using Formula]

⇒$\frac{1}{x^2}=\frac{7-4\sqrt{3}}{49-48}$

⇒$\frac{1}{x^2}=7-4\sqrt{3}$

Now adding $x^2 \:and\: \frac{1}{x^2}$

⇒$7{\cancel{+4\sqrt{3}}}+7{\cancel{-4\sqrt{3}}}$

⇒$\huge{\pink{\overbrace{\underbrace{Ans=14}}}}$

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Formulas Used :-

• $(a+b)^2=a^2+b^2+2ab$
• $(a+b)(a-b)=a^2-b^2$

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