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# Which of the following is NOT a root of the equation

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Joined: 11 Sep 2015
Posts: 4879
GMAT 1: 770 Q49 V46
Which of the following is NOT a root of the equation  [#permalink]

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25 Jul 2018, 06:13
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5
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Difficulty:

65% (hard)

Question Stats:

62% (02:36) correct 38% (02:35) wrong based on 137 sessions

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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

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Re: Which of the following is NOT a root of the equation  [#permalink]

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25 Jul 2018, 06:51
4
1
2
(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
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Re: Which of the following is NOT a root of the equation  [#permalink]

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25 Jul 2018, 06:55
1
GMATPrepNow wrote:
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

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
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4879
GMAT 1: 770 Q49 V46
Re: Which of the following is NOT a root of the equation  [#permalink]

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27 Jul 2018, 05:11
3
Top Contributor
GMATPrepNow wrote:
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

APPROACH 1: Plug in each answer choice to see which value does NOT satisfy the equation (slow but...)
For example, C) x = 3
We get: (3² + 3 - 20)² - 2(3² + 3 - 20) - 63 = 17
Evaluate: (-8)² - 2(-8) - 63 = 17
Simplify: 64 - 16 - 63 = 17
Works! So, x = 3 is a valid solution

Try D) x = 4
We get: (4² + 4 - 20)² - 2(4² + 4 - 20) - 63 = 17
Evaluate: (0)² - 2(0) - 63 = 17
Simplify: -63 = 17
Works! So, x = 4 is NOT a valid solution

------------------------------------------------

APPROACH 2: Let k = x² + x - 20
Now replace x² + x - 20 with k to get: k² - 2k - 63 = 17
Subtract 17 from both sides to get: k² - 2k - 80 = 0
Factor: (k - 10)(k + 8) = 0
So, either k = 10 or k = -8

Now replace k with x² + x - 20 to get:
x² + x - 20 = 10 and x² + x - 20 = -8

Take: x² + x - 20 = 10
Rearrange: x² + x - 30 = 0
Factor: (x + 6)(x - 5) = 0
Solutions: x = -6 and x = 5

Take: x² + x - 20 = -8
Rearrange: x² + x - 12 = 0
Factor: (x + 4)(x - 3) = 0
Solutions: x = -4 and x = 3

ALL SOLUTIONS: x = -6, 5, -4 and 3

-------------------------------------------------

APPROACH 3: Examine: x² + x - 20
Factor to get: (x + 5)(x - 4)
Notice that, when x = 4, (x + 5)(x - 4) = (4 + 5)(4 - 4) = 0

So, when x = 4, we get: (0)² - 2(0) - 63 = 17
When we simplify, we get: -63 = 17
In other words, x = 4 is NOT a solution to the original equation.

Cheers,
Brent
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Re: Which of the following is NOT a root of the equation  [#permalink]

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02 Sep 2019, 10:28
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Re: Which of the following is NOT a root of the equation   [#permalink] 02 Sep 2019, 10:28