Question

pleasee help me out.. it's basic maths question​

Answer 1

Given :

$\sqrt{20 + \sqrt{20 + \sqrt{20......} } } = x$

To find :

• The value of x

Solution :

It is given that :-

$\sqrt{20 + \sqrt{20 + \sqrt{20......} } } = x$

Therefore we can write the above expression as :-

$\sqrt{20 + x } = x$

Now after removing under root from LHS we get :-

${20 + x } = {x}^{2}$

or

${x}^{2} = {20 + x }$

${x}^{2} - x - 20 = 0$

Now we will solve the above expression by using middle term splitting .

(-5)×4 = -20

4-5 = -1

Therefore :-

${x}^{2} - 5x + 4x- 20 = 0$

[ From (4x-20 ) we can take out 4 as common and after that we get :- 4(x-5) ]

${x} (x - 5) + 4(x- 5) = 0$

taking out (x- 5) as common :-

$(x - 5) (x + 4) = 0$

Now , x-5=0 or x+4=0

Therefore x = 5 or x = -4

But x can't be negative . therefore x = 5

Answer :

x = 5