Question

If $x=\sqrt{7}+\sqrt{3}$ and $xy=4$, then $x^4+y^4=$

• A. $400$
• B. $368$
• C. $200$
• D. None of these

Answer 1

The correct option is B

$368$

Explanation for the correct option.

Step 1: Find the value of $y$.

$$\begin{gathered} xy=4 \\ \Rightarrow y=\frac{4}{x} \\ =\frac{4}{\sqrt{7}+\sqrt{3}}\times \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}} \\ =\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4} \\ =\sqrt{7}-\sqrt{3} \end{gathered}$$

$y=\sqrt{7}-\sqrt{3},x=\sqrt{7}+\sqrt{3}$,

Step 2: Find the value of $x^4+y^4$.

$\begin{array}{rcl}{x}^4+y^4& =& {\left(x^2+y^2\right)}^2-2x^2{y}^2\left[\mathrm{by}{a}^2+b^2={\left(a+b\right)}^2-2\mathrm{ab}\right]\\ & =& {\left[{\left(\sqrt{7}+\sqrt{3}\right)}^2+{\left(\sqrt{7}-\sqrt{3}\right)}^2\right]}^2-2{\left(xy\right)}^2\\ & =& {\left[7+3+2\left(\sqrt{7}·\sqrt{3}\right)+7+3-2\left(\sqrt{7}·\sqrt{3}\right)\right]}^2-32\\ & =& {\left(20\right)}^2-32\\ & =& 368\end{array}$

Hence, option B is correct.