If x - 10 = x^(1/2) + 10^(1/2), what is the value of x?

Question

If  x - 10 = \sqrt{x} + \sqrt{10} , what is the value of x?

(A)  1 + √10

(B)  10 + √10

(C)  11 + √10

(D)  10 + 2√10

(E)  11 + 2√10

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quizicalindian · Accepted Answer

It can be inferred from the prompt that this question wants to test use of:  a^2-b^2=(a-b)(a+b)

x-10 = (\sqrt{x})^2 - (\sqrt{10})^2 = (\sqrt{x}-\sqrt{10})(\sqrt{x}+\sqrt{10})

Since the question says,  x-10 = \sqrt{x}+\sqrt{10}  ; cross them out in the above equation

\sqrt{x}-\sqrt{10}=1

or  \sqrt{x}=1+\sqrt{10}

Then square this to get x :  1+10+2\sqrt{10}

Answer E

KarishmaB · Answer

If nothing comes to mind, simply approximate. x - 10 = \sqrt{x} + \sqrt{10} x - \sqrt{x} = 10 + \sqrt{10} sqrt(10) is slightly more than 3. So x - \sqrt{x} = 13 approx If x = 16, then x - \sqrt{x} = 16 - 4 = 12 If x = 17, then x - \sqrt{x} = 17 - 4.1 = 13 approx Hence x = 17 works. So we are looking for an option that gives about 17. Since sqrt(10) = 3.1 approx, only option (E) will give about 17. Answer (E) Video on estimations is here: https://youtu.be/4Wy7BrQrjkM

gmatophobia · Answer

x^2 - y^2 = (x+y)(x-y)

x - 10 = √x + √10

(\sqrt{x})^2 - (\sqrt{10})^2 = \sqrt{x} + \sqrt{10}

(\sqrt{x} + \sqrt{10})(\sqrt{x} - \sqrt{10}) = \sqrt{x} + \sqrt{10}

(\sqrt{x} + \sqrt{10})(\sqrt{x} - \sqrt{10}) -(\sqrt{x} + \sqrt{10}) = 0

(\sqrt{x} + \sqrt{10})  = 0

As  (\sqrt{x} + \sqrt{10})  is positive,  (\sqrt{x} - \sqrt{10}) - 1 = 0

(\sqrt{x} - \sqrt{10}) - 1 = 0

\sqrt{x} = \sqrt{10}+1

Squaring on both sides

x = 10 + 1 + 2\sqrt{10}

x = 11 + 2\sqrt{10}

Option E

Kinshook · Answer

Arghi 
Asked: If x - 10 = √x + √10, what is the value of x?

(√x + √10)(√x - √10) = √x + √10
(√x + √10) (√x - √10 - 1) = 0
√x = √10 + 1
x = (√10 + 1)^2 = 10 + 1 + 2√10 = 11 + 2√10

IMO E

Jake7Wimmer · Answer

Is there any other way to do this besides the algebra outlined above? There was no way I was recognizing that x - 10 should have been changed a^2−b^2.

Bunuel · Answer

That would be the best way in my mind. Alternatively, you could also substitute answers. Substituting E gives:
 
 (11 + 2√10) - 10 = \sqrt{x} + \sqrt{10}

1 + √10 = \sqrt{x} 
 
Squaring gives:
 
 1 + 2√10 + 10 = x

11 + 2√10 = x 
 
So, we got the value of x matching E.

MBAToronto2024 · Answer

KarishmaB  How better to solve it? I squared both sides as (x-sqrt(x))^2 = (10+sqrt(10))^2 and got smth similar to E.

KarishmaB · Answer

Squaring with x and root(x) terms is not a good idea. Upon squaring, we get x^2, x and root(x) terms. To remove root(x) we need to square again which will give us x^4 term which is just nasty! 
You need to do what gmatophobia did above if you are to use algebra. If that doesn't come to mind, then try to estimate the value as explained by me in the comment above.

rafin202 · Answer

Why can't i delete (√x + √10) from both side the 1st line?

rafin202 · Answer

Why can't i delete (√x + √10) from both side the 4th line?

Bunuel · Answer

Yes, since (√x + √10) cannot be 0, we can directly reduce (divide)

(√x + √10)(√x − √10) = √x + √10

by (√x + √10) to get

√x − √10 = 1.

We cannot reduce (divide) an equation by an expression if that expression could equal 0, as division by 0 is undefined, and doing so would result in losing a root. For example, if we had x^2 = x, reducing by x directly would give x = 1 but lose the root x = 0. Instead, we should proceed like this:

x^2 = x
x^2 − x = 0
x(x − 1) = 0

This gives both roots: x = 0 and x = 1.

In the given question, however, (√x + √10) cannot be 0 since it is the sum of non-negative (√x) and positive values (√10). Therefore, we can safely reduce (divide) the entire equation by (√x + √10) without losing any roots.

monarchme · Answer

man it was soo easy!! i just couldn't think of it

bumpbot · Answer

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If x - 10 = x^(1/2) + 10^(1/2), what is the value of x?

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If x - 10 = x^(1/2) + 10^(1/2), what is the value of x?

Prep Toolkit

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