Question

x=√5+√3, y=√5-√3 find the
${x}^{4 } - {y}^{4}$​

Answer 1

Answer:

hey mate here's your answer

${x}^{4} - {y}^{4} = ( {x}^{2} ) {}^{2} - ({y}^{2} ) {}^{2}$

$a {}^{2} - {b}^{2} = (a + b)(a - b)$

a=x² ; b= y²

$= ( {x}^{2} + {y}^{2} )( {x}^{2} - {y}^{2} )$

$( {x}^{2} + {y}^{2}) = ( \sqrt{5} + \sqrt{3} ) {}^{2} + ( \sqrt{5} - \sqrt{3} ) {}^{2}$

$( {x}^{2} - {y}^{2} ) = ( \sqrt{5} + \sqrt{3} ) {}^{2} - ( \sqrt{5} - \sqrt{3}) {}^{2}$

$( \sqrt{5} + \sqrt{3} ) {}^{2} = \sqrt{5} {}^{2} + \sqrt{3} {}^{2} + 2 \sqrt{15}$

$( \sqrt{5} - \sqrt{3} ) {}^{2} = \sqrt{5} { {}^{2} } + \sqrt{3} {}^{2} - 2 \sqrt{15}$

(x²+y²)=(5+3+2_/15)+(5+3-2_/15)

=16

(x²-y²)=(5+3+2_/15) - (5+3 - 2_/15)

4_/15

$= 16 \times 4 \sqrt{15} \\ = 64 \sqrt{15} \: \: \: \: \: (answer)$

may help dear.....✌❤

Answer 2

Given:

x=√5+√3 and y=√5-√3

To Find:

• Value of x⁴-y⁴

Solution:

We know that,

$\longrightarrow\mathrm{a^{2}-b^{2}=(a+b)(a-b)}$

$\longrightarrow\mathrm{a^{2}+b^{2}=(a+b)^{2}-2ab}$

So, firstly we will find the value of (x+y)²-2xy

$\mathrm{(x+y)^{2}-2xy}$

$\sf{(\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3})^{2}-2(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$

$\sf{(2\sqrt{5})^{2}-2((\sqrt{5})^{2}-(\sqrt{3})^{2})}$

$\sf{20-2(5-3)}$

$\sf{20-2(2)}$

$\sf{20-4}$

$\sf{16}$

Now,

$\longrightarrow\mathrm{x^{4}-y^{4}}$

$\longrightarrow\mathrm{(x^{2})^{2}-(y^{2})^{2}}$

$\longrightarrow\mathrm{(x^{2}+y^{2})(x^{2}-y^{2})}$

$\longrightarrow\mathrm{(x^{2}+y^{2})(x+y)(x-y)}$

$\longrightarrow\mathrm{((x+y)^{2}-2xy)(x+y)(x-y)}$

$\longrightarrow\sf{(16)(\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}-(\sqrt{5}-\sqrt{3}))}$

$\longrightarrow\sf{(16)(2\sqrt{5})(\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3})}$

$\longrightarrow\sf{(16)(2\sqrt{5})(2\sqrt{3})}$

$\longrightarrow\sf{64\sqrt{3\times5}}$

$\longrightarrow\sf\pink{64\sqrt{15}}$

Hence, the required value is  $\mathrm{\green{64\sqrt{15}}}$.