$\begin{aligned} \sqrt{x+15} + \sqrt x & = 15 \\ \sqrt{x+15} & = 15 - \sqrt x & \small \color{#3D99F6} \text{Squaring both sides.} \\ x+15 & = 225 - 30\sqrt x + x \\ 30\sqrt x & = 210 \\ \sqrt x & = 7 \\ \implies x & = \boxed{49} \end{aligned}$

By difference of squares , we know that $(a-b)(a+b) = a^2 - b^2$ .

In this case, let $a = \sqrt{x+15}, b = \sqrt x$ . We have

$\begin{aligned} \left( \sqrt{x+15} - \sqrt x \right) \left( \sqrt{x+15} + \sqrt x \right) &= & (x+15) - x \\ \left( \sqrt{x+15} - \sqrt x \right) \cdot 15 &=& 15 \\ \sqrt{x+15} - \sqrt x &=& 1 . \\ \end{aligned}$

Adding this equation to the original equation given shows

$\begin{aligned} \left( \sqrt{x+15} - \sqrt x \right) + \left( \sqrt{x+15} + \sqrt x \right) &=& 15 + 1 \\ 2\sqrt{x+15} &=& 16 \\ \sqrt{x+15} &=& 8 \\ x + 15 &=& 64 \\ x &=&\boxed{49}. \end{aligned}$

$\displaystyle (\sqrt{x + 15} + \sqrt{x})^2 = 15^2$

$\displaystyle x + 15 + 2\sqrt{x^2 + 15x} + x = 225$

$\displaystyle 2\sqrt{x^2 + 15x} = 210 - 2x$

$\displaystyle \sqrt{x^2 + 15x} = 105 - x$

$\displaystyle (\sqrt{x^2 + 15x})^2 = (105 - x)^2$

$\displaystyle x^2 + 15x = 105^2 - 210x + x^2$

$\displaystyle 225x = 11025$

$\displaystyle x = 49$

first time post: sequential squares of natural numbers increase by 2n+1. i just subtracted 1 from 15 and divided by 2 to get root x. root x =7 so x =49. these posts are hard to do on a phone

$\begin{aligned} & \sqrt{x+15} + \sqrt x &=& 15 \\ \implies & {(\sqrt{x+15})}^2 &=& {(15 - \sqrt x)}^2 \\ \implies & x+15 &=& 225 + x - 30 \sqrt x \\ \implies & \sqrt x &=& 7 \\ \implies & x &=& \boxed{49} \end{aligned}$

$\sqrt { x + 15} + \sqrt {x} = 15$

$\text {If } x \in \mathbb {R} \text { , then } x ≥ 0$

After squaring both sides and simplifying:

$2x + 15 + 2 \sqrt { x(x+15) } = 225$

$2 \sqrt { x(x+15) } = 210 - 2x$

$\sqrt { x^2 + 15x } = 105 - x$

After squaring again:

$x^2 + 15x = x^2 - 210x + 11025$

$225x = 11025$

$x = \boxed {49}$

Indeed:

$\sqrt { 49 + 15} + \sqrt {49} = 15$

$8 + 7 = 15$

Here's a simple solution. Since 15 is a whole number, we expect Sqrt(x) to at least be a rational number. It has to be the case that the square root of x is smaller than half of 15. First we assume that sqrt(x) is an integer, and so is sqrt(x+15). This requires both x and x+15 to be square numbers. By inspection, we can see that these conditions only hold for 49 and 64. If the square root of x is 7, then this becomes Sqrt(64) + Sqrt(49) = 8 + 7 = 15, as desired.

Let $a = \sqrt{x+15}$ and $b = \sqrt{x}$ . So, $a^2-b^2 = x+15-x = 15 \text{ and } a+b = 15,$ $a-b = \frac{a^2-b^2}{a+b} = \frac{15}{15} = 1 \text{ and } a+b = 15,$ $2b = (a+b)-(a-b) = 15-1 = 14,$ $b = 7,$ $x = b^2 = 7^2 = 49.$

This can be solved in a simple manner if one remembers the following

"n^2 = Sum of first n odd numbers"

ie.

1^2 = 1

2^2 = 1+3 = 1^2+3

3^2 = 1+3+5 = 2^2+5

and so on.

Here it is highly likely that √(x+15) and √x are both integers

15 is the 8th odd number and for (x+15) and x to be perfect squares x has to be the sum of first 7 odd numbers ie 7^2 = 49.

Just to be different.

$\sqrt{x+15}+\sqrt{x}=15$

Multiply by 15.

$15\sqrt{x}+15+15\sqrt{x}=225$

Subtract 15.

$15\sqrt{x}+15\sqrt{x}=210$

Simplify.

$30\sqrt{x}=210$

Divide by 30.

$\sqrt{x}=7$

Square.

$x=\boxed{49}$

Consider a right trangle, whose sides are $A = \sqrt{x}, ~B = \sqrt{15}, ~C = \sqrt{x + 15}$ . Then, the given equation is

$A + C = B^2$ (1)

But also, $B^2 = C^2 - A^2$ , from Pythagoras Theorem. Then,

$A + C = B^2 = C^2 - A^2 = (C + A)(C - A)$ , thereby,

$C - A = 1$ (2)

Therefore, substracting (2) from (1), one has that

$(A + C = 15) - (C - A = 1) = 2A = 14$

Hence $A = 7$ , but $A = \sqrt{x}$ , whence,

$x = 49$

None of the top answers took the slightly easier route. I don't know how to write solutions so they look pretty, but who cares...

Move the sqrt x to the other side

sqrt(x + 15) = 15 - sqrt(x)

Square both sides:

x + 15 = 225 - 30sqrt(x) + x

Cancel x, subtract 225 from both sides, divide by -30:

sqrt(x) = 7, x = 49

BOOM, NICE

$(\sqrt{x+15}-\sqrt{x})(\sqrt{x+15}+\sqrt{x})=15 (\sqrt{x+15}-\sqrt{x})$

$15 = 15 (\sqrt{x+15}-\sqrt{x})$

$\sqrt{x+15} - \sqrt{x} = 1$

taking the difference beetween the text and this equivalent equation we get:

$2 \sqrt{x} = 14$

$7 = \sqrt{x}$

First reasoning: I suppose $x+15$ and x are perfect square. So $m^2-n^2=15$ and $(m-n)(m+n)=3\cdot5$ so that m=4 and n=1 but this couple does'nt solve the equation, otherwise $(m-n)(m+n)=15 \cdot 1$ and we get m=8, n=7 and this couple is right. So that $x=n^2=49$ . And I was lucky when I supposed x could be an Integer.

You can solve this by simply plugging-in. Sum of square root of two numbers have to be 15. One is that number and another is the number + 15. That reduces the possible candidates to 64 and 49 from a list of whole squares candidates ( 1,4,9,16,25,36..). So x has to be 49

From the above equation it can be said that 15 is the summation of two full number square integer whose difference is 15 .
now let's just find out the square numbers
1²=1;
2²=4; difference=3.
3²=9; difference=5.
4²=16; difference=7.
5²=25; difference=9.
6²=36; difference=11.
7²=49; difference=13.
8²=64; difference=15.
So the answer should be 49. (no mathematical calculation is needed.)

$\sqrt{x+15}+\sqrt{x}=15$

$\textbf{Solution}$

$\sqrt{x} = 15-\sqrt{x+15}$

Square both sides, expand, and simplify ...

$(\sqrt{x})^2 = (15-\sqrt{x+15})^2$

$x=(15)^2-2\cdot15\cdot\sqrt{x+15}+(\sqrt{x+15})^2$

${\cancel{x}}=225-30\sqrt{x+15}+{\cancel{x}}+15$

Try to isolate $x$ ...

$30\sqrt{x+15}=225+15$

$30\sqrt{x+15}=240$

$\sqrt{x+15}=\frac{240}{30}$

Again square both sides to get rid of the unwanted root ...

$x+15=8^2$

And you end up with ...

$x=64-15$

$\therefore \boxed{x=49}$

$\textbf{Proof}$

Plug in the value and you get...

$\sqrt{49+15}+\sqrt{49}$

which yields...

$8 + 7 = 15$ .

$Finding \ a \ general \ solution, \\ let \ 15=y \\ y=\sqrt { x+y } +\sqrt { x } (1) \\ squaring \ both \ sides \\ { y }^{ 2 }=y+2x+2\sqrt { x(x+y) } \\ { y }^{ 2 }=y+2\sqrt { x } (\sqrt { x } +\sqrt { x+y } ) \\ from \ (1) \\ { y }^{ 2 }=y+2\sqrt { x } (y) \\ (y-1)[y-(1+2\sqrt { x } )]=0 \\ y=1 \ as \ y=15 \ this \ equation \ is \ useless \\. therefore \ y-(1+2\sqrt { x } )=0 \\ \sqrt { x } =\frac { y-1 }{ 2 } \\ x={ (\frac { y-1 }{ 2 } ) }^{ 2 } \\ at \ y=15 , \\ x=49 \\$

$\sqrt{x+15} + \sqrt{x} = 15$

Subtract $\sqrt{x}$ from both sides, we obtain

$\sqrt{x+15} = 15 - \sqrt{x}$

Square both sides of the equation,

$(\sqrt{x+15})^2 = (15 - \sqrt{x})^2$

$x + 15 = 15^2 - 30\sqrt{x} + x$ , from here $x$ and $x$ cancels out

$15 + 225 - 30\sqrt{x}$

Simplifying further, we have

$30\sqrt{x} = 210$

Divide both sides by $30$ then square both sides,

$x = 49$

Do you know that this question is used on an advert for Brilliant?

Even $x$ is a real number. We could expect that $\sqrt{x}$ and $\sqrt{x+15}$ are positive integers for the first try.

Since $2\sqrt{x} < \sqrt{x+15} + \sqrt{x} = 15$ , we have $\sqrt{x} < 7.5$ .

A quick test with $\sqrt{x} = 7$ give us the correct answer.

Mathematica

NSolve[Sqrt[x + 15] + Sqrt[x] == 15, x]

{{x -> 49.}}

Do it mentally.

Try using the square numbers, obviously the ones greater than 15. for eg- * 25, 36,49,64,81 * Remember, your x has to be a square number, to be get easily rooted, and so does (x+15). Now choose a square number from the list, that when you add 15 to it gives you another square number. For eg: 49+15= 64, now replace (x+15) with 64 and x with 49. √64 + √49 = 15

We can see that both x+15 and x has to be perfect squares. Now in 2 or 3 hit and trials we know the answer is 49. there are straight methods for everything but i believe in some cases tricks might save time and solving equations.

$\sqrt{x+15} + \sqrt{x} = 15 \space\space\space\space\space (1)$

$\frac{(\sqrt{x+15} + \sqrt{x})(\sqrt{x+15} - \sqrt{x})}{\sqrt{x+15} - \sqrt{x}} = 15$

$\frac{15}{\sqrt{x+15} - \sqrt{x}} = 15$

$\sqrt{x+15} - \sqrt{x} = 1 \space\space\space\space\space (2)$

$(1) - (2) \space : \space 2\sqrt{x} = 14$

$\sqrt{x} = 7$

$x = 49$

√x+15 + √x = 15

√x+15 = 15 - √x

(√x+15 = 15 - √x)^2

x + 15 = (15-√)(15-√x)

x + 15 = 225 - 15√x - 15√x + x

-30√x = 210

√x = -7

x = 49

(\sqrt{x+15}+\sqrt{x})^{2}=15^{2} x+15+2\sqrt{(x+15)*x}+15=225 2\sqrt{(x^{2}+15x}=225-15-2x (2\sqrt{(x^{2}+15x})^{2}=(210-2x)^{2} 4(x^{2}+15x)=44100-840x+4x^{2} 900x=44100 x=\frac{44100}{900} x=49

Here x must be moderately large. The square roots of x + 15 and x must be very similar. e.g. 8 and 7. Try x = 49. No need for any complicated maths here!

x=a^2, a^2+15=(a+1)^2

so a =7, x=49.

Premise: The solution is easy. That means that both $\sqrt x$ and $\sqrt{ x + 15 }$ are integers. Probably consecutive. As everyone knows $(n+1)^2 - n^2 = 2n + 1$ , therefore if $2n+1 = 15$ , then $n = 7$ . Indeed $7+8 = 15$ ; this leads to a solution $x = n^2 = 49$ . By a mental manipulation of the identity, and squaring, one can observe that it has a unique solution.

By conjugation method , without any squaring both side.

$\sqrt{x+15} + \sqrt{x} = 15$

${\sqrt{x+15} - \sqrt{x}} = 1$

Subtract both equation

$\begin{aligned} 2\sqrt{x} = 14 \\ \sqrt{x} = 7 \\ x = 49 \end{aligned}$

$\sqrt{x+15} + \sqrt{x} = 15$

$(\sqrt{x+15} + \sqrt{x}) (\frac{\sqrt{x+15} - \sqrt{x}}{\sqrt{x+15} - \sqrt{x}}$ ) = 15 ＜－ multiply Left Hand Side with its conjugate

$\frac{(x+15) - (x)}{\sqrt{x+15} - \sqrt{x}} = 15$

$\frac{15}{\sqrt{x+15} - \sqrt{x}} = 15$

$\frac{1}{\sqrt{x+15} - \sqrt{x}} = 1$

${\sqrt{x+15} - \sqrt{x}} = 1$

simplify the equation by transferring one radical at the right side of the equation. then square both sides. then simplify again and then use the quadratic equation in getting the value of x. x=49

Reason: √(x+15)=15-√x Now, square both sides of the equation X+15=225-30√x+x 15=225-30√x 30√x=210 √x=7, Therefore, x=49

$\sqrt{(x+15)}+\sqrt{(x)}=15$

we need to find 2 squares, whose square roots add up to 15, 15 apart

we find 49 and 64 with 7+8=15

$\sqrt{(x+15)}+\sqrt{(x)}=\sqrt{(49+15)}+\sqrt{(49)}=15$

$\sqrt{(x)}=\sqrt{(49)}=>x=49$

The "simpler" is the key word. I´ve tried the first perfect natural squares (1,4,9,16,25,36,49,64) for diference of 15. then 49 is the correct answer.

Using Newton Raphson Method: $x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$ we find the value of $x$ in six iterations.

The difference between two consecutive square numbers is an odd number and we now that sqrt(x+15) and sqrt(x) are even numbers so x is a perfect square and x+15 is also a perfect square. (if x wasnt integer the sum would never be integer). So you only have two choices 1 (+3+5+7)= 16 and 49+15=64... well just try 1 and 49 and see that 49 is right

conventional maths is okay. lets try aptitude.

assumption 1: the numbers are perfect square.

think of x such that its a perfect square and adding 15 results in a perfect square too.

we have to test it on 1,4,9,16,25,36,49,64.... (which are perfect squares) 1 and 49 satisfies the condition and 49 satisfies the equation.

√(x+15) + √x=15 √(x+15) + x - 2 √(x+15) √x = 225 (by squaring both sides) or, 2x=210+2 √(x+15) √x or,x=105+√(x+15)*√x or,(x+105)² = x² + 15x or,225x = 11025 so, x = 49

Both √x and x have to be a natural number, so the only options for x are: 1,4,9,16,25,36,49,64,... Knowing that, you can instantly find that x = 49 without any complex calculation, just trial-and-error basis.

Same as Chew-Seong Cheong

• éléver au carrée les deux membres de l'équation;
• mettre les racines dans un membres et les nombres sans racines dans un autre membre;
• éléver au carrée les deux membres et là on obtient une équation sans racine carré de degré 1 après réduction de terme;
• résoudre l'équation de degré 1 et obtenir x=49 après simplification de la fraction obtenue.