Math, asked by Anonymous, 1 year ago

Solve basic algebra pls guys

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Answered by darky1

put x=11 then
$\sqrt{11 + 14} < 11 + 2 \\ \sqrt{25} < 13 \\ 5 < 13 \\ mark \: both \: answer \: brainliest$

darky1: mark as brainliest

Anonymous: how x is equal to 11

AmeyaShri123: what?

darky1: you have to put the values and try if you take x equal 10 then you will unable to solve it

Anonymous: i want detailed answer edit ur answer

darky1: we have to make is a square number which is 25

darky1: put the value of x from 1 to 10 it will not satisfy,the only way to solve this type of problems is to try and put different values starting fromm 1

darky1: or it can be done by squaring on both sides

Answered by abhi569

$\sqrt{ x + 14} < x + 2 \\ \\ \\$

Square on both sides,

$x + 14 < {x}^{2} + 4 + 2x \\ \\ => 0 < {x}^{2} + 4x - x + 4 - 14 \\ \\ => 0 < {x}^{2} + 3x - 10$

=> 0 < x² + 5x - 2x - 10

=> 0 < x(x + 5) - 2(x + 5)

=> 0 < (x + 5)(x - 2)

x = 2 or - 5

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