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08 Mar 2025, 15:05
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EnriqueDandolo
Hey Bunuel, I was wondering if you could please explain step by step in more detail how did you apply the algebraic identity (x^2 - y^2)=(x+y)(x-y) to the fraction
Square root(2)*[Square root(x) + Square root(y)] /(x-y).
Many thanks in advance.
I am confused because after applying the same algebraic identity I got as a result the following:
Square root(2)*[Square root(x) + Square root(y)]*(x+y)/(x^2 - y^2)
Square root(2)*[Square root(x) + Square root(y)] /(x-y).
Many thanks in advance.
I am confused because after applying the same algebraic identity I got as a result the following:
Square root(2)*[Square root(x) + Square root(y)]*(x+y)/(x^2 - y^2)
Bunuel
3. If \(x>y>0\) then what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)
\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\) --> factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).
(1) \(x+y=4+2\sqrt{xy}\) --> \(x-2\sqrt{xy}+y=4\) --> \((\sqrt{x}-\sqrt{y})^2=4\) --> \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y>0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.
(2) \(x-y=9\). Not sufficient.
Answer: A.
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)
\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\) --> factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).
(1) \(x+y=4+2\sqrt{xy}\) --> \(x-2\sqrt{xy}+y=4\) --> \((\sqrt{x}-\sqrt{y})^2=4\) --> \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y>0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.
(2) \(x-y=9\). Not sufficient.
Answer: A.
We are applying \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator"
In the denominator we have x - y, which can be written as \(x-y=(\sqrt{x})^2-(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})\).
Hope it's clear.
09 Mar 2025, 11:05
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Hi experts Bunuel KarishmaB, I was wondering if you could please help me point what I am doing wrong here. To evaluate statement 2, I would actually do the following to the stimulus:
Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:
Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.
After simplifying, the expression would turn into:
x+y+z/cube root(xyz).
Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)
Thus, the result of the expression will always be 0.
Many thanks in advance for your help.
Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:
Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.
After simplifying, the expression would turn into:
x+y+z/cube root(xyz).
Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)
Thus, the result of the expression will always be 0.
Many thanks in advance for your help.
Bunuel
9. If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\) --> infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.
(2) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: \(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.
Must know for the GMAT: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B.
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\) --> infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.
(2) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: \(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.
Must know for the GMAT: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B.
09 Mar 2025, 11:32
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EnriqueDandolo
Hi experts Bunuel KarishmaB, I was wondering if you could please help me point what I am doing wrong here. To evaluate statement 2, I would actually do the following to the stimulus:
Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:
Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.
After simplifying, the expression would turn into:
x+y+z/cube root(xyz).
Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)
Thus, the result of the expression will always be 0.
Many thanks in advance for your help.
Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:
Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.
After simplifying, the expression would turn into:
x+y+z/cube root(xyz).
Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)
Thus, the result of the expression will always be 0.
Many thanks in advance for your help.
Bunuel
9. If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\) --> infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.
(2) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: \(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.
Must know for the GMAT: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B.
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\) --> infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.
(2) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: \(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.
Must know for the GMAT: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B.
The point is \(\sqrt{x^3+y^3+z^3}\) does not equal to x^3+y^3+z^3. You'd be right if it were (x + y + x)^3 instead.
