The largest among the numbers sqrt{7}-sqrt{5} | vedclass

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Question

(B) To compare numbers of the form $\sqrt{x+2}-\sqrt{x}$,we rationalize the expression by multiplying and dividing by the conjugate $\sqrt{x+2}+\sqrt{

Vedclass · Accepted Answer

$\sqrt{5}-\sqrt{3}$

Vedclass · Answer

(B) To compare numbers of the form $\sqrt{x+2}-\sqrt{x}$,we rationalize the expression by multiplying and dividing by the conjugate $\sqrt{x+2}+\sqrt{x}$.$\sqrt{x+2}-\sqrt{x} = \frac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \frac{(x+2)-x}{\sqrt{x+2}+\sqrt{x}} = \frac{2}{\sqrt{x+2}+\sqrt{x}}$.As $x$ increases,the denominator $\sqrt{x+2}+\sqrt{x}$ increases,which means the value of the fraction $\frac{2}{\sqrt{x+2}+\sqrt{x}}$ decreases.Comparing the given numbers:$1$. For $\sqrt{5}-\sqrt{3}$,$x=3$,value $= \frac{2}{\sqrt{5}+\sqrt{3}}$.$2$. For $\sqrt{7}-\sqrt{5}$,$x=5$,value $= \frac{2}{\sqrt{7}+\sqrt{5}}$.$3$. For $\sqrt{9}-\sqrt{7}$,$x=7$,value $= \frac{2}{\sqrt{9}+\sqrt{7}}$.$4$. For $\sqrt{11}-\sqrt{9}$,$x=9$,value $= \frac{2}{\sqrt{11}+\sqrt{9}}$.Since the denominator is smallest for $\sqrt{5}-\sqrt{3}$,the value of $\sqrt{5}-\sqrt{3}$ is the largest.

The largest among the numbers $\sqrt{7}-\sqrt{5}$,$\sqrt{5}-\sqrt{3}$,$\sqrt{9}-\sqrt{7}$,and $\sqrt{11}-\sqrt{9}$ is:

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