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D
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Difficulty:
65%
(hard)
Question Stats:
55% (01:43) correct
45%
(01:17)
wrong
based on 800
sessions
History
Date
Time
Result
Not Attempted Yet
Which of the following equations is true for all positive values of \(x\) and \(y\)?
A) \(\sqrt{x}+\sqrt{y}=\sqrt{x+y}\)
B) \(\sqrt{x^4y^{16}} =x^2y^4\)
C) \((x\sqrt{y})(y\sqrt{x})=x^2y^2\)
D) \(y\sqrt{x}+y\sqrt{x}=\sqrt{4xy^2}\)
E) \((x^y)(y^y)=(xy)^{2y}\)
A) \(\sqrt{x}+\sqrt{y}=\sqrt{x+y}\)
B) \(\sqrt{x^4y^{16}} =x^2y^4\)
C) \((x\sqrt{y})(y\sqrt{x})=x^2y^2\)
D) \(y\sqrt{x}+y\sqrt{x}=\sqrt{4xy^2}\)
E) \((x^y)(y^y)=(xy)^{2y}\)
26 May 2018, 08:32
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Princ
Properties of exponents and roots: Eliminate A, B, and E
Incorrect arithmetic: Eliminate C
A) \(\sqrt{x}+\sqrt{y}=\sqrt{x+y}\)
Exponents and roots do not distribute over addition and subtraction
Let \(x = 16, y = 4\)
\(\sqrt{16}+\sqrt{4}\neq\sqrt{20}\)
\(4 + 2 = 6\) but \(\sqrt{20} = 2\sqrt{5}\)
B) \(\sqrt{x^4y^{16}} =x^2y^4\)
\(\sqrt{x^4y^{16}}\) \(=x^2y^8\)
Exponents and roots:
\(\sqrt{y^{16}}=(y^{16})^{\frac{1}{2}}=\) \(y^8\)
--NOT \(y^4\)
C) \((x\sqrt{y})(y\sqrt{x})=x^2y^2\)
\((x\sqrt{y})(y\sqrt{x})=x*y*\sqrt{y}*\sqrt{x}\)
\(=xy\sqrt{xy}\)
E) \((x^y)(y^y)=(xy)^{2y}\)
Different base, same exponent: multiply the bases, keep the exponent
The exponent should be \(y\), not \(2y\)
\(a^{x}b^{x}=(ab)^{x}\)
\((x^y)(y^y)=(xy)^{y}\)
Let \(x = 3\) and \(y = 2\)
Answer (E) as it stands: \((3^2)(2^2)=(6)^{4}=\) \(1,296\)
Corrected:
\((3^2)(2^2)=(3*2)^{2}=6^2=36\)
\((3^2)(2^2)=(9)(4)=36\)
By POE, ANSWER D
D) \(y\sqrt{x}+y\sqrt{x}=\sqrt{4xy^2}\)
\(y\sqrt{x}+y\sqrt{x}=2y\sqrt{x}\)
\((2y\sqrt{x})^2= 4y^2x\)
\(\sqrt{(2y\sqrt{x})^2}=\sqrt{4y^2x}\)
\(y\sqrt{x}+y\sqrt{x}=\sqrt{4xy^2}\)
Correct
ANSWER D
EDIT: I changed my approach to D (no effect on answer).
CAMANISHPARMAR had already chosen the wiser route, but I didn't notice.
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