Symbiosis National Aptitude Test SNAP 2019 Paper

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Question : 68 of 110

Marks: +1, -0

Given

(√{x+4}+√{x-10})^2/{(x+4)-(x-10)}

=

5/2

, find the value of x

${264}/{20}$
${263}/{20}$
${265}/{20}$
${266}/{20}$

Solution:

(√{x+4}+√{x-10})^2/{(x+4)-(x-10)}

=

5/2

or

{(x+4)+(x-10)+2√{x^2-6x-40}}/{14}

=

5/2

or

3x+2√{x^2-6x-40}

= 35 + 6
or (2x - 41) =

-2√{x^2-6x-40}

or

(2x - 41)^2

= 4(

x^2

- 6x - 40)
or

4x^2

+1681-164x =

4x^2

- 24x -160
or 140x = 1841 or x =

{263}/{20}

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