✓5+√3÷√5-√3 = a+b✓15 find a and …

{tex}\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+\sqrt{15} b{/tex}

Rationalizing means the process of rationalizing the denominator, by rewriting the denominator for make the denominator into rational numbers.

Rationalizing the above equation so that,  

{tex}\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=a+\sqrt{15} b{/tex}

[By multiplying {tex}\sqrt{5}+\sqrt{3}{/tex} in both numerator and denominator of the LHS of equation]

{tex}\frac{(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}-\sqrt{3})^{2}}=a+\sqrt{15} b{/tex}

[Since {tex}(a+b)^{2}=a^{2}+b^{2}+2 a b{/tex}

{tex}(a+b)(a-b)=a^{2}+b^{2}]{/tex}

{tex}\frac{5+3+2 \sqrt{5} \times \sqrt{3}}{(5-3)}=a+\sqrt{15} b{/tex}

{tex}\frac{8+2 \sqrt{15}}{(2)}=a+\sqrt{15} b{/tex}

{tex}4+\sqrt{15}=a+\sqrt{15} b{/tex}

{tex}4+\sqrt{15} \times 1=a+\sqrt{15} b{/tex}

By Comparing both side of the equations, then we get:

a= 4;b = 1