Question

if y= √√7+7+√8+2√7 -√7 find value of y

Answer 1

We see that,

y=√√7+7+√8+2√7−√7

Here,

8+2√7=7+1+2√7=(√7+1)2

On solving for y , we get

y=√√7+7+√(√7+1)2−√7

y=√√7+7+(√7+1)−√7

y=√2√7+7+1−√7

Using identity, we get

y=√(√7+1)2−√7

y=(√7+1)−√7

∴y=1

Answer 2

The value of y = 1

Correct question : If $\displaystyle \sf{y = \sqrt{ \sqrt{7} + 7 + \sqrt{8 + 2 \sqrt{7} } } - \sqrt{7} }$ find the value of y

Given :

$\displaystyle \sf{y = \sqrt{ \sqrt{7} + 7 + \sqrt{8 + 2 \sqrt{7} } } - \sqrt{7} }$

To find :

The value of y

Solution :

Step 1 of 3 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{y = \sqrt{ \sqrt{7} + 7 + \sqrt{8 + 2 \sqrt{7} } } - \sqrt{7} }$

Step 2 of 3 :

Find the square root of 8 + 2√7

$\displaystyle \sf{ 8 + 2 \sqrt{7} }$

$\displaystyle \sf{ = {( \sqrt{7} )}^{2} + {1}^{2} + 2 \times \sqrt{7} \times 1}$

$\displaystyle \sf{ = {( \sqrt{7} + 1)}^{2} }$

$\displaystyle \sf{ \therefore \: \: \sqrt{8 + 2 \sqrt{7} } = \sqrt{7} + 1} \: \: \: (taking \: positive \: sign)$

Step 3 of 3 :

Find the value of y

$\displaystyle \sf{y = \sqrt{ \sqrt{7} + 7 + \sqrt{8 + 2 \sqrt{7} } } - \sqrt{7} }$

$\displaystyle \sf{ \implies y = \sqrt{ \sqrt{7} + 7 + ( \sqrt{7} + 1)} - \sqrt{7} }$

$\displaystyle \sf{ \implies y = \sqrt{8 + 2 \sqrt{7}} - \sqrt{7} }$

$\displaystyle \sf{ \implies y = \sqrt{7} + 1 - \sqrt{7} }$

$\displaystyle \sf{ \implies y = 1}$

Hence the value of y = 1

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