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29 Nov 2012, 12:02
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:51) correct
32%
(01:54)
wrong
based on 366
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of x?
(1) \(3x^2 + 8x + \frac{16}{3} = 0\)
(2) \(15x^2 + \frac{80}{3} = -40x\)
(1) \(3x^2 + 8x + \frac{16}{3} = 0\)
(2) \(15x^2 + \frac{80}{3} = -40x\)
29 Nov 2012, 13:54
Kudos
Bookmarks
What is the value of x?
(1) 3x^2 + 8x + 16/3 = 0 --> multiply by 3: \(9x^2+24x+16=0\) --> \((3x+4)^2=0\) --> \(x=-4/3\). Sufficient.
(2) 15x^2 + 80/3 = -40x --> multiply by 3/5: \(9x^2+16=-24x\) --> \(9x^2+24x+16=0\) --> \((3x+4)^2=0\) --> \(x=-4/3\). Sufficient.
Answer: D.
(1) 3x^2 + 8x + 16/3 = 0 --> multiply by 3: \(9x^2+24x+16=0\) --> \((3x+4)^2=0\) --> \(x=-4/3\). Sufficient.
(2) 15x^2 + 80/3 = -40x --> multiply by 3/5: \(9x^2+16=-24x\) --> \(9x^2+24x+16=0\) --> \((3x+4)^2=0\) --> \(x=-4/3\). Sufficient.
Answer: D.
29 Nov 2012, 15:49
Kudos
Bookmarks
0k, that's a perfect square.
Otherwise, I reduced to the standard quadratic form: ax^2 + bx + c and factorize.
9x^2 + 24x + 16 = 0
9x^2 + 12x + 12x + 16 = 0
3x(3x+4) + 4(3x+4)=0
(3x + 4)^2 = 0
Or,
x^2 + (8/3)x + (4/3)^2 = 0 [(8/3)^2 = 4*(1)(16/9)]
(x + 4/3)^2 = 0....x= -4/3
Otherwise, I reduced to the standard quadratic form: ax^2 + bx + c and factorize.
9x^2 + 24x + 16 = 0
9x^2 + 12x + 12x + 16 = 0
3x(3x+4) + 4(3x+4)=0
(3x + 4)^2 = 0
Or,
x^2 + (8/3)x + (4/3)^2 = 0 [(8/3)^2 = 4*(1)(16/9)]
(x + 4/3)^2 = 0....x= -4/3
