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If x is a real number and x + x^(1/2) = 1, which of the following is

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yrozenblum
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If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   06 Nov 2023, 10:28
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If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)
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If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   06 Nov 2023, 11:44
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yrozenblum wrote:
If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)


Assume \(\sqrt{x} = y\)

\(x + \sqrt{x} = 1\)

\(y^2 + y= 1\)

\(y^2 + y - 1 = 0\)

The roots of a quadratic equation represented by \(ax^2 + bx + c\) is given by \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Hence, the roots of the equation, \(y^2 + y - 1 = 0\), is given by

\(\frac{-1\pm\sqrt{1-(4*-1)}}{2}\)

\(\frac{-1\pm\sqrt{1+4}}{2}\)

\(\frac{-1\pm\sqrt{5}}{2}\)

As, \(y = \sqrt{x}\) → The value will be non negative

\(\sqrt{x} =\frac{-1 + \sqrt{5}}{2}\)

Option A .
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MartyMurray
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Re: If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   30 Apr 2024, 13:56
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Alternative Approach

If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)

Since \(x + \sqrt{x} = 1\), we know that x is positive because we can take the square root of x.

We also know that \(x < 1\) and that \(\sqrt{x} < 1\).

So, we can eliminate (B) and (D) since \(\sqrt{5} > 1\).

We can also eliminate (C) because it will be negative since \(\sqrt{5} < 3\).

So, we're down to (A) and (E).

We can estimate the values of (A) and (E) by considering that \(\sqrt{5} \) must be between \(2\) and \(3\).

So, (A) is around \(\frac{1}{2}(2.25 -1) = \frac{5}{8} \).

(E) is around \(\frac{1}{2}(3 - 2.25) = \frac{3}{8} \).

So, \((A) > \frac{1}{2}\), and \((E) < \frac{1}{2}\).

Now, since \(x < 1\), \(\sqrt{x} > x\).

So, for \(x + \sqrt{x} = 1\), it  must be that \(\sqrt{x} > 1/2\).

Thus, (E) is out, and only (A) can work.

Correct answer: A­­
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Re: If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   25 May 2024, 00:21
What is wrong with this approach? 

x + x^1/2 = 1 --> x^1/2 = 1 - x

x^1/2 = 1 - x --> (x^1/2)^2 = (1 - x)^2

(x^1/2)^2 = (1 - x)^2 --> x = 1 + x^2 - 2x

x = 1 + x^2 - 2x --> x^2 - 3x + 1 = 0

Then using the quadratic equation we get the roots  3 + (5^1/2)/2 and 3 - (5^1/2)/2 and since x must be postive then Answer D. 3 + (5^1/2)/2?
gmatophobia wrote:
yrozenblum wrote:
If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)

Assume \(\sqrt{x} = y\)

\(x + \sqrt{x} = 1\)

\(y^2 + y= 1\)

\(y^2 + y - 1 = 0\)

The roots of a quadratic equation represented by \(ax^2 + bx + c\) is given by \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Hence, the roots of the equation, \(y^2 + y - 1 = 0\), is given by

\(\frac{-1\pm\sqrt{1-(4*-1)}}{2}\)

\(\frac{-1\pm\sqrt{1+4}}{2}\)

\(\frac{-1\pm\sqrt{5}}{2}\)

As, \(y = \sqrt{x}\) → The value will be non negative

\(\sqrt{x} =\frac{-1 + \sqrt{5}}{2}\)

Option A .

­
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Re: If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   25 May 2024, 01:51
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kevinhirose wrote:
What is wrong with this approach? 

x + x^1/2 = 1 --> x^1/2 = 1 - x

x^1/2 = 1 - x --> (x^1/2)^2 = (1 - x)^2

(x^1/2)^2 = (1 - x)^2 --> x = 1 + x^2 - 2x

x = 1 + x^2 - 2x --> x^2 - 3x + 1 = 0

Then using the quadratic equation we get the roots  3 + (5^1/2)/2 and 3 - (5^1/2)/2 and since x must be postive then Answer D. 3 + (5^1/2)/2?
gmatophobia wrote:
yrozenblum wrote:
If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)

Assume \(\sqrt{x} = y\)

\(x + \sqrt{x} = 1\)

\(y^2 + y= 1\)

\(y^2 + y - 1 = 0\)

The roots of a quadratic equation represented by \(ax^2 + bx + c\) is given by \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Hence, the roots of the equation, \(y^2 + y - 1 = 0\), is given by

\(\frac{-1\pm\sqrt{1-(4*-1)}}{2}\)

\(\frac{-1\pm\sqrt{1+4}}{2}\)

\(\frac{-1\pm\sqrt{5}}{2}\)

As, \(y = \sqrt{x}\) → The value will be non negative

\(\sqrt{x} =\frac{-1 + \sqrt{5}}{2}\)

Option A .

­

­
Several points:

1. The roots of \(x^2 - 3x + 1 = 0\) are \(\frac{3 - \sqrt{5}}{2}\approx 0.4\) and \(\frac{3 + \sqrt{5}}{2}\approx 2.6\). Both are positive, however, \(\frac{3 + \sqrt{5}}{2}\approx 2.6\) is not a valid root of \(x + \sqrt{x} = 1\): \(2.6 + \sqrt{2.6} > 1\). (When you squared you get that extra invalid root.)

2. We are asked to find the value of \(\sqrt{x}\), not x. So, if we substitute \(x=\frac{3 - \sqrt{5}}{2}\) into \(x + \sqrt{x} = 1\), we get \(\frac{3 - \sqrt{5}}{2} + \sqrt{x} = 1\), which yields \(\sqrt{x} = 1 - \left(\frac{3 - \sqrt{5}}{2}\right) = \frac{\sqrt{5} - 1}{2}\).

Hope it's clear.­
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If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink] If x is a real number and x + x^(1/2) = 1, which of the following is   25 May 2024, 04:08
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Because all mathematical explanations are already discussed so I am discussing a Purely Logical approach here

Given: x+√x = 1
Step-1: x = 1-√x

RULES: now, this is possible only when value is between 0 and 1
Also, the square root of any value between 0 and 1 gets bigget than the value itself e.g. √0.25 = 0.5


i.e. if √x = 0.8 then x = 0.64 but the sum of x and √x is greater than 1 lets take another approximation
if √x = 0.6 then x = 0.36 and the sum of x and √x in this case is 0.96 so very close to 1

so the correct option should be close to 0.6, Let's check options now,

A. \(\frac{1}{2}(\sqrt{5}-1)\) = (1/2)*(2.3-1) = 1.3/2 = 0.65 (Pretty close so possible answer)

B. \(\frac{1}{2}(\sqrt{5}+1)\) = (1/2)*(2.3+1) = 3.3/2 = 1.65 (Not possible)

C. \(\frac{1}{2}(\sqrt{5}-3)\) = (1/2)*(2.3-3) = -0.7/2 = Negative value (Not possible)

D. \(\frac{1}{2}(\sqrt{5}+3)\) = (1/2)*(2.3+3) = 5.3/2 = 2.65 (Not possible)

E. \(\frac{1}{2}(3-\sqrt{5})\) = (1/2)*(3-2.3) = 0.7/2 = 0.35 (not close to 0.6 hence Not possible)

Answer: Option A

P.S.: if you think that this approach doesn't work in real exam then I must say you are mistaken. This works wonderfully and flawlessly

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yrozenblum wrote:
If \(x\) is a real number and \(x + \sqrt{x} = 1\), which of the following is the value of \(\sqrt{x}\)?

A. \(\frac{1}{2}(\sqrt{5}-1)\)

B. \(\frac{1}{2}(\sqrt{5}+1)\)

C. \(\frac{1}{2}(\sqrt{5}-3)\)

D. \(\frac{1}{2}(\sqrt{5}+3)\)

E. \(\frac{1}{2}(3-\sqrt{5})\)

­
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If x is a real number and x + x^(1/2) = 1, which of the following is [#permalink]
25 May 2024, 04:08
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