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25 Jul 2018, 07:13
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:33) correct
33%
(02:51)
wrong
based on 419
sessions
History
Date
Time
Result
Not Attempted Yet
Which of the following is NOT a root of the equation (x2 + x - 20)2 - 2(x2 + x - 20) - 63 = 17
A) -6
B) -4
C) 3
D) 4
E) 5
*Note: there are at least 3 different solutions possible
A) -6
B) -4
C) 3
D) 4
E) 5
*Note: there are at least 3 different solutions possible
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25 Jul 2018, 07:51
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(x² + x - 20)² - 2(x² + x - 20) - 63 = 17
Let (x² + x - 20) = m
=> \(m^2 -2m -63 = 17\)
=> \(m^2 -2m -80 = 0\)
=> \(m^2 - 10m + 8m - 80 = 0\)
=> \(m(m-10) + 8(m-10) = 0\)
=> \((m + 8)(m - 10) = 0\)
=> m = -8 or m = 10
Case 1 when m = -8
x² + x - 20 = -8
=> \(x^2 + x -12 = 0\)
=> \(x^2 + 4x - 3x - 12 = 0\)
=> \(x(x+4) - 3(x+4) = 0\)
=> \((x-3)(x+4) = 0\)
=> x = 3 or x = -4
Case 2 when m = 10
x² + x - 20 = 10
=> \(x^2 + x -30 = 0\)
=> \(x^2 + 6x - 5x - 30 = 0\)
=> \(x(x+6) - 5(x+6) = 0\)
=> \((x+6)(x-5) = 0\)
=> x = -6 or x = 5
=> x can be 3,-4,-6,5
Hence option D as x cannot be 4
Let (x² + x - 20) = m
=> \(m^2 -2m -63 = 17\)
=> \(m^2 -2m -80 = 0\)
=> \(m^2 - 10m + 8m - 80 = 0\)
=> \(m(m-10) + 8(m-10) = 0\)
=> \((m + 8)(m - 10) = 0\)
=> m = -8 or m = 10
Case 1 when m = -8
x² + x - 20 = -8
=> \(x^2 + x -12 = 0\)
=> \(x^2 + 4x - 3x - 12 = 0\)
=> \(x(x+4) - 3(x+4) = 0\)
=> \((x-3)(x+4) = 0\)
=> x = 3 or x = -4
Case 2 when m = 10
x² + x - 20 = 10
=> \(x^2 + x -30 = 0\)
=> \(x^2 + 6x - 5x - 30 = 0\)
=> \(x(x+6) - 5(x+6) = 0\)
=> \((x+6)(x-5) = 0\)
=> x = -6 or x = 5
=> x can be 3,-4,-6,5
Hence option D as x cannot be 4
General Discussion
25 Jul 2018, 07:55
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GMATPrepNow
OA: D
1st Method
\((x^2+x-20)^2 - 2(x^2+x-20) -63 = 17\)
\((x^2+5x-4x-20)^2 - 2(x^2+5x-4x-20) = 17+63\)
\([(x+5)(x-4)]^2 -2(x+5)(x-4)=80\)
If we put \(x=4\) , \(L.H.S=0\) and \(R.H.S=80\)
\(L.H.S\neq{R.H.S}\)
So 4 is NOT a root of the equation (x² + x - 20)² - 2(x² + x - 20) - 63 = 17
2nd Method
\((x^2+x-20)^2 - 2(x^2+x-20) -63 = 17\)
\((x^2+x-20)^2 - 2(x^2+x-20) = 17+63\)
Adding 1 to both L.H.S and R.H.S, we get
\((x^2+x-20)^2 - 2(x^2+x-20)+1 = 81\)
\((x^2+x-20 -1)^2=9^2\) (Using \(a^2+b^2-2ab =(a-b)^2\))
\((x^2+x-21)=±9\)
Case 1: \(x^2+x-21 = +9\)
\(x^2+x-30=0\)
\((x+6)(x-5)=0, x=-6,5\)
Case 2: \(x^2+x-21 = -9\)
\(x^2+x-12=0\)
\((x+4)(x-3)=0, x=-4,3\)
Root of expression in questionstem are \(-6,5,-4,3\)
Only option D is not there
