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What is the value of x^2
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07 Sep 2021, 07:17
07 Sep 2021, 07:17
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If \(\sqrt{3-x}\) \(= \sqrt{x} + 3\) , then \(x^2=\)
(A) 1
(B) 3
(C) 2 − 3x
(D) 3x − 1
(E) 3x − 9
(A) 1
(B) 3
(C) 2 − 3x
(D) 3x − 1
(E) 3x − 9
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: What is the value of x^2
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07 Sep 2021, 09:57
07 Sep 2021, 09:57
\(3-x=x+9+6\sqrt{x}\)
\(6+2x=-6\sqrt{x}\)
\(3+x=-3\sqrt{x}\)
\(9+x^2+6x=9x\)
\(x^2=3x-9\)
answer is E
\(6+2x=-6\sqrt{x}\)
\(3+x=-3\sqrt{x}\)
\(9+x^2+6x=9x\)
\(x^2=3x-9\)
answer is E
Carcass wrote:
If \(\sqrt{3-x}\) \(= \sqrt{x} + 3\) , then \(x^2=\)
(A) 1
(B) 3
(C) 2 − 3x
(D) 3x − 1
(E) 3x − 9
(A) 1
(B) 3
(C) 2 − 3x
(D) 3x − 1
(E) 3x − 9
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Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: What is the value of x^2
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03 Sep 2022, 04:41
03 Sep 2022, 04:41
Given that \(\sqrt{3-x}\) \(= \sqrt{x} + 3\) and we need to find the value of \(x^2=\)
\(\sqrt{3-x}\) \(= \sqrt{x} + 3\)
Squaring both the sides we get
\((\sqrt{3-x})^2\) \(= (\sqrt{x} + 3)^2\)
=> 3 - x = \((\sqrt{x})^2 - 2*\sqrt{x}*3 + 3^2\)
=> 3 - x = x - 6\(\sqrt{x}\) + 9
=> 3 - 2x - 9 = - 6\(\sqrt{x}\)
=> -2x - 6 = -6\(\sqrt{x}\)
Dividing both sides by -2
=> x + 3 = 3\(\sqrt{x}\)
Squaring both the sides we get
=> \((x+3)^2\) = \((3\sqrt{x})^2\)
=> \(x^2 + 2*3*x + 3^2\) = 9x
=> \(x^2\) = 9x - 6x - 9 = 0
=> \(x^2\) = 3x − 9
So, Answer will be E
Hope it helps!
\(\sqrt{3-x}\) \(= \sqrt{x} + 3\)
Squaring both the sides we get
\((\sqrt{3-x})^2\) \(= (\sqrt{x} + 3)^2\)
=> 3 - x = \((\sqrt{x})^2 - 2*\sqrt{x}*3 + 3^2\)
=> 3 - x = x - 6\(\sqrt{x}\) + 9
=> 3 - 2x - 9 = - 6\(\sqrt{x}\)
=> -2x - 6 = -6\(\sqrt{x}\)
Dividing both sides by -2
=> x + 3 = 3\(\sqrt{x}\)
Squaring both the sides we get
=> \((x+3)^2\) = \((3\sqrt{x})^2\)
=> \(x^2 + 2*3*x + 3^2\) = 9x
=> \(x^2\) = 9x - 6x - 9 = 0
=> \(x^2\) = 3x − 9
So, Answer will be E
Hope it helps!
