How to solve sqrt(x + 15) + sqrt(x) = 15 ?

[richp123]

I have been away from my math pursuits for some time. I don't remember how to solve the following equation for x

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

Any suggestions are appreciated how to approach the solution for this equation.

 

[richp123]

I vaguely remember the needing to use a quadratic equation and squaring both sides. I guess now my question is about using the quadratic formula. It's slowly coming back to me.

 

[Greg]

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

Rearrange:

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

Square both sides:

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

Subtract $x$ from both sides:

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

Rearrange:

$$30\sqrt{x}=210$$

Divide both sides by 30:

$$\sqrt{x}=7$$

$$x=49$$

Does that help?

 

[kaliprasad]

I have been away from my math pursuits for some time. I don't remember how to solve the following equation for x

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

Any suggestions are appreciated how to approach the solution for this equation.

alternatively

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

we know $(x+15)- x = 15$ (identity)
or $(\sqrt{x + 15} + \sqrt{x}))(\sqrt{x + 15}- \sqrt{x}) = 15\cdots(2)$

deviding (2) by (1)

$\sqrt{x + 15} - \sqrt{x} = 1\cdots(3)$

add (1) and (3) to get

$2\sqrt{x + 15} = 16$

or $\sqrt{x + 15} = 8$ or $x+15=64$ or $x=49$

 

[AshKetchum]

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

Rearrange:

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

Square both sides:

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

Subtract $x$ from both sides:

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

Rearrange:

$$30\sqrt{x}=210$$

Divide both sides by 30:

$$\sqrt{x}=7$$

$$x=49$$

Does that help?

Could you help me understand where the 30 comes from in $$x+15=225-30\sqrt{x}+x$$

 

[Greg]

Hi AshKetchum and welcome to MHB! :)

$$(15-\sqrt x)^2=(15-\sqrt x)(15-\sqrt x)=15\cdot15-15\sqrt x-15\sqrt x+x=225-30\sqrt x+x$$

Does that help?

 

[AshKetchum]

Hi AshKetchum and welcome to MHB! :)

$$(15-\sqrt x)^2=(15-\sqrt x)(15-\sqrt x)=15\cdot15-15\sqrt x-15\sqrt x+x=225-30\sqrt x+x$$

Does that help?

Perfectly. Thank you!

 

[HOI]

In general, (a+ b)^2= (a+ b)(a+ b)= a(a+ b)+ b(a+ b)= a^2+ ab+ ba+ b^2= a^2+ 2ab+ b^2.

Notice that this depends upon the fact that, for a and b numbers, ab= ba. If we were working in an algebraic system in which multiplication was not "commutative" that would not be true.

 

[ozdigennaro]

Assuming that x is must be an integer.
It's easiest to simply start testing integers.
And they must have an integer square root. That makes it pretty easy.
Just guess!

And there's another numerical clue.

The difference between n squared and (n+1) squared = n + n +1
So 15 is the giveaway. 7 + 8 = 15
So 7 squared is 49.
You can do it All In Your Head. No algebra necessary. That algebra warps your mind.

Find the easy way!

In fact, if you look at this truth: "difference between n squared and (n+1) squared = n + n +1"
You'll get the answer in milliseconds.

 

[blueecho]

I'm now stuck on this problem, I understand the other solutions in this thread but why can't you square all parts of the equation at the outset?

sqrt(x-15)^2 + sqrt(x)^2 = 15^2

PS. just pulled up a mathjax tutorial since sqrt(x) is atrocious to read.

 

[HOI]

Because, as you have been told repeatedly here, the square of $\sqrt{x- 15}+ \sqrt{x}$ is NOT $x- 15+ x$.

To square $a+ b$ you multiply $(a+ b)(a+ b)$. Using the "distributive law", that is $a(a+ b)+ b(a+ b)= a^2+ ab+ ba+ b^2= a^2+ 2ab+ b^2$, NOT $a^2+b^2$.

If that is not sufficiently convincing, $(3+ 4)^2= 7^2= 49$ NOT $3^2+ 4^2= 9+ 16= 25$. It IS $3^2+ 2(3)(4)+ 4^2= 9+ 24+ 16= 25+ 24= 49$.

 

[blueecho]

Got it, thanks