There's only one answer right?

Algebra Level 2

$\large \sqrt{x + \sqrt{ y + \sqrt{x + \sqrt{y + \sqrt{ x + \sqrt{y + \ldots}}} } } } = 7$

If $y=2$ , then what is the value of $x$ ?

The answer is 46.

4 solutions

Sravanth C.
May 17, 2015

According to the question, $\large \sqrt{x + \sqrt{ y + \sqrt{x + \sqrt{y + \sqrt{ x + \sqrt{y + \ldots}}} } } } = 7$

Substituting, $y=2$ in the equation, we get, $\large \sqrt{x + \sqrt{ 2 + \sqrt{x + \sqrt{2 + \sqrt{ x + \sqrt{2 + \ldots}}} } } } = 7$

Or, $\sqrt{x + \sqrt{ 2 + 7 } } = 7$

Squaring both the sides, $x+\sqrt{9}=7^{2}$ $x+3=49$ $x=49-3$ $x=46$

Or, by taking $\sqrt{9}$ as negitve, we get $x=52$ as the new result, because the problem poser has given $46$ as the answer

Moderator note:

Can you prove that the nested function converges?

since $\sqrt{9}$ = +3 or -3, I think there should be two solutions. 46 and 52

Santosh Sadan - 6 years ago

I also got 52 as the other answer but i think that the square root sign is conventionally used to give positive result.

Nihar Mahajan - 6 years ago

No, both the signs in a squareroot are correct according to maths. And its wrong to neglect the negetive result.

In some cases we neglect one result in a squareroot to take it as a measure of physical quantity. But it doesnt make sense that the other esult was wrong. It actually how we resolve the meaning of how the other can be right. (Got this answer actually from a lecture of a yale university professor who actually proved that why in some cases we get two answers as a part of squareroot. He proved both will always be correct. )

Vivek Vijayan - 6 years ago

@Vivek Vijayan – Thanks. I have edited it.

Sravanth C. - 6 years ago

@Vivek Vijayan – When you are taking radical (i.e square root) , then $\sqrt{x}= |x|$

Nihar Mahajan - 6 years ago

Thanks Nihar, I have edited it.

Sravanth C. - 6 years ago

Thanks for informing, I have edited it.

Sravanth C. - 6 years ago

In response to challenge master: Sorry @Calvin Lin sir, I'd come across, radicals of the form $\sqrt{a+\sqrt {a+ \sqrt{a + . . . .}}}$ , $\sqrt{a+b\sqrt {a+b \sqrt{a +b. . . . }}}$ , $\sqrt{a-b\sqrt {a-b \sqrt{a -b . . . . }}}$ and few more.

Can you help me in proving it please?

Sravanth C. - 6 years ago

just wanna ask how do you answer this? sqrt(2+sqrt(2+sqrt(2+.....)))=?

Caeo Tan - 6 years ago

Let, $\Huge \sqrt{2 \sqrt{2 \sqrt{2 . . . . }}}=x$

Or, $\huge \sqrt{2x}=x$

$\huge x^{2}=2x$

Therefore, $\huge x=\boxed{2}$

Sravanth C. - 6 years ago

it's sqrt(2 + sqrt(2 + sqrt(2 + ....)))=?

Caeo Tan - 6 years ago

@Caeo Tan – Ohh, okay, here the solution.

$\Huge \sqrt{2+ \sqrt{2+ \sqrt{2 +. . . . }}}=x$

$\Huge \sqrt{2 + x}=x$

$\Huge 2+x=x^{2}$

$\Huge x^{2}-x-2=0$

Therefore, $\huge x$ has two values,

i.e, $\huge \dfrac{1+\sqrt{9}}{2}$ or, $\huge \dfrac{1-\sqrt{9}}{2}$

Of which the second one is negative so we can ignore it.

Sravanth C. - 6 years ago

@Sravanth C. – what if it's sqrt(2 + sqrt(2 - sqrt(2 + sqrt(2 - ....))))=?

Caeo Tan - 6 years ago

@Caeo Tan – $\Huge \sqrt{2+ \sqrt{2- \sqrt{2 +\sqrt{2-. . . . }}}}=x$

$\Huge \sqrt{2 + \sqrt{2 - x}}=x$

$\Huge 2 + \sqrt{2 - x}=x^{2}$

$\Huge x^{2}-2=\sqrt{2-x}$

$\Huge [x^{2}-2]^{2}=2-x$

$\huge x^{4}+4-4x^{2}=2-x$

$\huge x^{4}+2+x-4x^{2}=0$

Solving the equation above we get four values for $\huge x$ ,

i.e, $\huge \boxed{\boxed{ \dfrac{1+\sqrt{5}}{2}}}$ , $\huge \boxed{\boxed{1}}$ , $\huge\boxed{\boxed{-2}}$ and $\huge \boxed{\boxed{\dfrac{1-\sqrt{5}}{2}}}$

$\huge\ddot\smile$

Sravanth C. - 6 years ago

@Caeo Tan – Please see that there was a mistake in calculation, in the last question.

Sravanth C. - 6 years ago

@Sravanth C. – What does these negative solutions signify in geometry?

Ashwin Gopal - 6 years ago

How u got 7 by putting y=2...

Deepansh Jindal - 6 years ago

Please go through the question once again, it's hidden somewhere $\huge\ddot\smile$

Sravanth C. - 6 years ago

No there should be only 1solution X=49-√9 is a 1 degree equation so only 1 soln. Consider x=√9 so x=3 and not -3 ....hard to absorb ...but it is the fact

prakhar mishra - 5 years, 11 months ago

Chris M.
Jun 1, 2015

Given this equation:

$\large \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}}}}=7$

We can square both sides of the equation: $\large x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}}}=7^2=49$

Subtract x from both sides:

$\large \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}}}=49-x$

Square both sides again:

$\large y + \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}}=(49-x)^2$

Refer back to our original equation (the position of the ellipses does not matter, as both continue forever and converge to 7): $\large \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}}}}=7$

Substitute 7 for the nested square root: $\displaystyle \large y + 7=(49-x)^2$

This is a good time to substitute given $y=2$ : $\displaystyle 2 + 7=(49-x)^2$ $\displaystyle 9=(49-x)^2$ $\displaystyle \pm3=49-x$ $\displaystyle 49-x=-3 \: \:OR \: \:49-x=3$ $\displaystyle x=49+3 \: \: OR \: \: x=49-3$ $\displaystyle x=52 \: \: OR \: \: x=46$

Ador El
May 18, 2015

y=2. So, sqrt(x + sqrt(2 + sqrt(x +...))) = 7

we see that sqrt(x+...) inside the expression on the left is 7.

So, sqrt(x + (sqrt(2 + 7))) = 7

doing algebra

x + sqrt(9) = 49 x + 3 = 49 x = 46

Good solution @Ador El , but it would be better if you had used ${\LaTeX}$ ,learn more about it here .

Sravanth C. - 6 years ago

Caeo Tan
May 18, 2015

sqrt(x+sqrt(2+7))=7||sqrt(x+3)=7|| x+3=49||x=46

Good solution @Caeo Tan sir, but it would have been better if you'd used ${\LaTeX}$ learn more about it here .

Sravanth C. - 6 years ago

There's only one answer right?

The answer is 46.

4 solutions

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