Question

If x=√28+5√12 and y=√37+10√12, What I'd the value of (x-y/(x+y)

Answer 1

Step-by-step explanation:

x=√28+5√12

y=√37+10√12

x + y =√28+5√12 +(√37+10√12)

=√28+15√12 + √37

x-y =√28+5√12-(√37+10√12)

=√28-5√12-√37

so x-y/(x+y) =(√28-5√12-√37)/ (√28+15√12+√37)

Answer 2

Step-by-step explanation:

Given:If

$x = \sqrt{28 + 5\sqrt{12} } \: \: and \: \: y = \sqrt{ 37 + 10\sqrt{12} }$

To find: Find the value of

$\frac{x - y}{x + y}$

Solution:

Tip: Convert the terms under square root into complete whole square.

Step 1: Convert x in complete square

Rearrange the terms

$x = \sqrt{28 + 5 \sqrt{12} }$

or

$x=\sqrt{28 + 5 \sqrt{4 \times 3} }$

or

$x=\sqrt{28 + 2\times 5\sqrt{3} }$

or

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

because

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

or

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

or

$\bold{\red{x = 5 + \sqrt{3}}}$

Step 2: Convert y into complete square

$y = \sqrt{ 37 + 10 \sqrt{12} }$

or

$y = \sqrt{37 + 2 \times 5 \sqrt{12} }$

or

$y = \sqrt{( {5)}^{2} + 2 \times 5 \sqrt{12} + ( { \sqrt{12} )}^{2} }$

or

$y = \sqrt{( {5 + \sqrt{12}) }^{2} }$

or

$\bold{\green{y = 5 + \sqrt{12} }}$

Step 3: Put value of x and y

$\frac{x - y}{x + y}$

$\frac{x - y}{x + y}=\frac{5 + \sqrt{3} - 5 - \sqrt{12} }{5 + \sqrt{3} + 5 + \sqrt{12} }$

or

$\frac{x - y}{x + y}=\frac{ \sqrt{3} - \sqrt{12} }{10 + \sqrt{3} + \sqrt{12} }$

or

$\frac{x - y}{x + y}=\frac{ \sqrt{3} - 2 \sqrt{3} }{10 + \sqrt{3} + 2 \sqrt{3} }$

or

$\frac{x - y}{x + y} = \frac{ - \sqrt{3} }{10 + 3 \sqrt{3} }$

Step 4: Rationalize the denominator

$\frac{x - y}{x + y} = \frac{ - \sqrt{3} }{10 + 3 \sqrt{3} } \times \frac{10 - 3 \sqrt{3} }{10 - 3 \sqrt{3} }$

apply identity a²-b²=(a+b)(a-b) in the denominator

$\frac{x - y}{x + y} = \frac{ - 10 \sqrt{3} + 9 }{( {10)}^{2} - ( {3 \sqrt{3}) }^{2} }$

or

$\frac{x - y}{x + y} = \frac{ - 10 \sqrt{3} + 9 }{100 - 27 }$

or

$\frac{x - y}{x + y} = \frac{9 - 10 \sqrt{3} }{73 }$

Final answer:

$\bold{\pink{\frac{x - y}{x + y} = \frac{9 - 10 \sqrt{3} }{73 }}}$

Hope it helps you.

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