GMAT CLUB OLYMPICS: If k 0 and (k - 1/k)

Question

If k ≠ 0 and  \sqrt{k - \frac{1}{k}} + \sqrt{1 - \frac{1}{k}} = k , then what is the value of k ?

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

B.  \frac{\sqrt{5}-1}{2}

C.  \sqrt{5}-1

D.  \frac{1+\sqrt{5}}{2}

E.  \sqrt{5}+1

M37-33

Bunuel · Accepted Answer

Official Solution:

If  k ≠ 0  and  \sqrt{k - \frac{1}{k}} + \sqrt{1 - \frac{1}{k}} = k , then what is the value of  k  ?

A.  \frac{1-\sqrt{5}}{2} 
B.  \frac{\sqrt{5}-1}{2} 
C.  \sqrt{5}-1 
D.  \frac{1+\sqrt{5}}{2} 
E.  \sqrt{5}+1

Step 1: Define variables 
 
Let  \sqrt{k - \frac{1}{k}}=x  and  \sqrt{1 - \frac{1}{k}}=y . We will use these variables to simplify the given equation.
 
 Step 2: Write two equations 
 
From the given equation, we have:
 
 x+y=k ;
 
Multiply both sides by  (x-y)  to obtain  (x-y)(x+y)=k(x-y) . 
 
 Step 3: Simplify the equation 
 
 x^2-y^2=k(x-y) ; 
 
 (k - \frac{1}{k})-(1 - \frac{1}{k})=k(x-y) ; 
 
 k-1=k(x-y) ;
 
 x-y=\frac{k-1}{k} . 
 
 Step 4: Solve the system of equations 
 
We now have two equations:  x+y=k  and  x-y=\frac{k-1}{k} .
 
Add these two equations:  2x=k+\frac{k-1}{k}=k+1-\frac{1}{k} . 
 
 Step 5: Substitute and solve for  x  
 
Substitute  \sqrt{k - \frac{1}{k}}=x  into  2x=(k-\frac{1}{k})+1 :
 
 2x=x^2+1 ;
 
 (x-1)^2=0 ;
 
 x=1 .
 
 Step 6: Solve for  k  
 
Now we have  \sqrt{k - \frac{1}{k}}=x=1 ;
 
Squaring both sides:  k - \frac{1}{k}=1 ;
 
 k^2 - k-1=0 ;
 
This gives us two possible values for  k :  k = \frac{1-\sqrt{5}}{2}  or  k = \frac{1+\sqrt{5}}{2} .
 
 Step 7: Determine the valid value of  k  
 
The first root is negative, which is not valid because the sum of two non-negative values ( \sqrt{k - \frac{1}{k}}  and  \sqrt{1 - \frac{1}{k}} ) cannot equal a negative number ( k ). Therefore, the valid value for  k  is  k = \frac{1+\sqrt{5}}{2} .

Answer: D

Berdiyor · Answer

Hello everyone

Here is my solution 
The answer is D

Kinshook · Answer

Asked: If k ≠ 0 and \sqrt{k - \frac{1}{k}} + \sqrt{1 - \frac{1}{k}} = k , then what is the value of k ? \sqrt{k - \frac{1}{k}} + \sqrt{1 - \frac{1}{k}} = k \sqrt{k - \frac{1}{k}} = k - \sqrt{1 - \frac{1}{k}} Squaring both sides k - 1/k = k^2 + (1-1/k) - 2k\sqrt{1 - \frac{1}{k}} k^2 - k + 1 = 2k\sqrt{1 - \frac{1}{k}} (k^2 - k + 1)^2 = 4k^2 (1-1/k) = 4k^2 - 4k = 4k(k-1) (k(k-1) + 1)^2 = 4k(k-1) Let k(k-1) = t (t+1)^2 = 4t t^2 + 2t + 1 = 4t t^2 - 2t + 1 = 0 (t - 1)^2 = 0 t = 1 k(k-1) = 1 k^2 - k - 1 = 0 k = \frac{(1 +- \sqrt{1 + 4})}{2} k = \frac{(1 + - \sqrt{5})}{2} Since (k - 1/k) >=0 & (1- 1/k) >=0 (1- 1/k) >=0 (k - 1)/k >=0 k >=1 or k<0 (k - 1/k) >=0 (k^2 - 1)/k >=0 (k-1)(k+1)/k >=0 k>=1 or -1<=k<0 Combining k>=1 or -1<=k<0 k = \frac{1-\sqrt{5}}{2} = -.61 k = \frac{1+\sqrt{5}}{2} = 1.61>1 Since summation of 2 square roots is a positive value, k = \frac{1+\sqrt{5}}{2} = 1.61>1 IMO D

cs15811581 · Answer

The best way is to substitute the values.

A.  \frac{1-\sqrt{5}}{2}  will be a negative value. Discard

B.  \frac{\sqrt{5}-1}{2}=1.2/2=0.6 
 \sqrt{0.6 - \frac{1}{0.6}} + \sqrt{1 - \frac{1}{0.6}} = 0.6 
Discard as the second root comes out to be negative.

C.  \sqrt{5}-1=1.2 
 \sqrt{1.2 - \frac{1}{1.2}} + \sqrt{1 - \frac{1}{1.2}} = 1.2 
 \sqrt{1.2 - 0.83 }+ \sqrt{1 -0.83} = k 
 \sqrt{0.36 }+ \sqrt{0.16} = 1.2 
 0.6+0.4=1.2 ….Discard

D.  \frac{1+\sqrt{5}}{2} 
 \sqrt{1.6 - \frac{1}{1.6}} + \sqrt{1 - \frac{1}{1.6}} = 1.6 
 \sqrt{1.6 - 0.625}+ \sqrt{1 -0.625} = 1.6 
 \sqrt{1}+ \sqrt{0.4} = 1.6 
 1+0.62=1.6 ….
Almost the same
Correct

E.  \sqrt{5}+1

D

sivatx2 · Answer

https://gmatclub.com/forum/download/file.php?mode=view&id=63218 SquareRootProblem.png

bumpbot · Answer

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GMAT CLUB OLYMPICS: If k 0 and (k - 1/k)

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