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29 Aug 2022, 07:57
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B
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Difficulty:
95%
(hard)
Question Stats:
27% (01:55) correct
73%
(01:49)
wrong
based on 95
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of x?
(\(1\)) \(\sqrt{2x^2 - 4x - 9} = \sqrt{x^2 - 3x + 3}\)
(\(2\)) \(\sqrt{2x^2 + x - 16} = \sqrt{x^2 + 2x - 4}\)
(\(1\)) \(\sqrt{2x^2 - 4x - 9} = \sqrt{x^2 - 3x + 3}\)
(\(2\)) \(\sqrt{2x^2 + x - 16} = \sqrt{x^2 + 2x - 4}\)
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29 Aug 2022, 09:56
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What is the value of x?
1. \(\sqrt{2x^2 - 4x - 9} = \sqrt{x^2 - 3x + 3}\)
Square both sides
\(2x^2-4x-9=x^2-3x+3…….x^2-x-12=0…..(x-4)(x+3)=0\)
So, two values. But wait!! what if one of them is NOT possible.
When x=4, \(x^2-3x+3=4^2-3*4+3=7\)…ok
When x=-3, \(x^2-3x+3=(-3)^2-3*(-3)+3=21\)
2. \(\sqrt{2x^2 + x - 16} = \sqrt{x^2 + 2x - 4}\)
Square both sides
\(2x^2+x-16=x^2+2x-4…….x^2-x-12=0…..(x-4)(x+3)=0\)
So, two values. But wait!! what if one of them is NOT possible.
When x=4, \(x^2+2x-4=4^2+2*4-4=20\)…ok
When x=-3, \(x^2+2x-4=(-3)^2+2*(-3)-4=-1\)….Not possible
So, only possibility is that x is 4.
Sufficient
B
1. \(\sqrt{2x^2 - 4x - 9} = \sqrt{x^2 - 3x + 3}\)
Square both sides
\(2x^2-4x-9=x^2-3x+3…….x^2-x-12=0…..(x-4)(x+3)=0\)
So, two values. But wait!! what if one of them is NOT possible.
When x=4, \(x^2-3x+3=4^2-3*4+3=7\)…ok
When x=-3, \(x^2-3x+3=(-3)^2-3*(-3)+3=21\)
2. \(\sqrt{2x^2 + x - 16} = \sqrt{x^2 + 2x - 4}\)
Square both sides
\(2x^2+x-16=x^2+2x-4…….x^2-x-12=0…..(x-4)(x+3)=0\)
So, two values. But wait!! what if one of them is NOT possible.
When x=4, \(x^2+2x-4=4^2+2*4-4=20\)…ok
When x=-3, \(x^2+2x-4=(-3)^2+2*(-3)-4=-1\)….Not possible
So, only possibility is that x is 4.
Sufficient
B
30 Aug 2022, 07:55
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BrentGMATPrepNow
Target question: What is the value of x?
Statement 1: √(2x² - 4x - 9) = √(x² - 3x + 3)
Square both sides: 2x² - 4x - 9 = x² - 3x + 3
Rearrange: x² - x - 12 = 0
Factor: (x - 4)(x + 3) = 0, which means x = 4 or x = -3
Test the possible solutions!
Plug x = 4 into the original equation to get: √(2(4²) - 4(4) - 9) = √(4² - 3(4) + 3)
Simplify: √7 = √7. Works! So, x can equal 4
Plug x = -3 into the original equation to get: √(2(-3)² - 4(-3) - 9) = √((-3)² - 3(-3) + 3)
Simplify: √21 = √21. Works! So, x can equal -3
Since there are two possible values of x, statement 1 is not sufficient.
Statement 2: √(2x² + x - 16) = √(x² + 2x - 4)
Square both sides: 2x² + x - 16 = x² + 2x - 4
Rearrange: x² - x - 12 = 0
This is the same equation we derived for statement 1, which means x = 4 or x = -3
Test the possible solutions!
Plug x = 4 into the original equation to get: √(2(4²) + 4 - 16) = √(4² + 2(4) - 4)
Simplify: √20 = √20. Works! So, x can equal 4
Plug x = -3 into the original equation to get: √(2(-3)² + (-3) - 16) = √((-3)² + 2(-3) - 4)
Simplify: √-1 = √-1.
Note: √-1 is not a real number.
Since the GMAT deals only with real numbers, x = -3 is not a valid solution.
So, it must be the case that x = 4
So, statement 2 is sufficient.
Answer: B
