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# M18-35

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Math Expert
Joined: 02 Sep 2009
Posts: 50042

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16 Sep 2014, 01:04
00:00

Difficulty:

45% (medium)

Question Stats:

59% (01:08) correct 41% (01:14) wrong based on 215 sessions

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If set $$M$$ consists of the root(s) of equation $$2-x^2 = (x-2)^2$$, what is the range of set $$M$$?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

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Math Expert
Joined: 02 Sep 2009
Posts: 50042

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16 Sep 2014, 01:05
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Official Solution:

If set $$M$$ consists of the root(s) of equation $$2-x^2 = (x-2)^2$$, what is the range of set $$M$$?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

$$2-x^2 = (x-2)^2$$;

$$2-x^2=x^2-4x+4$$;

$$x^2-2x+1=0$$;

$$(x-1)^2=0$$;

$$x=1$$. So, set $$M$$ consists of only one element.

The range of a single element set is 0.

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Concentration: Accounting
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25 May 2015, 14:43
Is there anyway that someone can show me how they got the answer of 0? I can get this far:

2-x^2 = (X-2)^2

2-X^2 = (X-2) (X-2)

2-X^ 2 = X^2-4X+4

What to do after this. I don't understand the break down on the free GMAT text. It's not clicking at this moment.
Math Expert
Joined: 02 Sep 2009
Posts: 50042

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26 May 2015, 06:32
whitdiva23 wrote:
Is there anyway that someone can show me how they got the answer of 0? I can get this far:

2-x^2 = (X-2)^2

2-X^2 = (X-2) (X-2)

2-X^ 2 = X^2-4X+4

What to do after this. I don't understand the break down on the free GMAT text. It's not clicking at this moment.

$$2 - x^ 2 = x^2 - 4x + 4$$;

Re-arrange: $$2x^2 - 4x +2 = 0$$;

Reduce by 2: $$x^2 -2x + 1 = 0$$, which is the same as $$(x−1)^2=0$$, so $$x=1$$.
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10 Feb 2017, 09:47
Hi Bunuel

I think

$$(x−1)^2$$=0 will give 2 equal values for X(1,1) not single value . So range will be 0.

Is this correct ?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 50042

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10 Feb 2017, 10:00
pranjal123 wrote:
Hi Bunuel

I think

$$(x−1)^2$$=0 will give 2 equal values for X(1,1) not single value . So range will be 0.

Is this correct ?

Thanks

1 and 1 is just one root.
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Joined: 03 May 2014
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16 Nov 2017, 08:56
1
Bunuel wrote:
whitdiva23 wrote:
Is there anyway that someone can show me how they got the answer of 0? I can get this far:

2-x^2 = (X-2)^2

2-X^2 = (X-2) (X-2)

2-X^ 2 = X^2-4X+4

What to do after this. I don't understand the break down on the free GMAT text. It's not clicking at this moment.

$$2 - x^ 2 = x^2 - 4x + 4$$;

Re-arrange: $$2x^2 - 4x +2 = 0$$;

Reduce by 2: $$x^2 -2x + 1 = 0$$, which is the same as $$(x−1)^2=0$$, so $$x=1$$.

How would one know when to reduce the equation as opposed to trying to solve a quadratic equation ax^2 +bx + c when a>1. For example factoring out the 2x^2?
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Joined: 20 Mar 2018
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23 May 2018, 11:39
What about x=0 as another root.
I agree that x=1 is one solution to the squared quadratic but then there are two roots to the solution right, x=0 and x=1. Hence, i thought the range should be 1.

Math Expert
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23 May 2018, 21:25
nshivapu wrote:
What about x=0 as another root.
I agree that x=1 is one solution to the squared quadratic but then there are two roots to the solution right, x=0 and x=1. Hence, i thought the range should be 1.

x = 0 does NOT satisfy $$2-x^2 = (x-2)^2$$.

LHS = 2 - 0 = 2, while RHS = (0 - 2)^2 = 4.
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11 Jul 2018, 10:02
$$x=1$$. So, set $$M$$ consists of only one element.

The range of a single element set is 0.

I didn't understand from here on. What's the logic? I know that range is the output, but how do we determine it?
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Math Expert
Joined: 02 Sep 2009
Posts: 50042

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11 Jul 2018, 11:14
Tanisha_shasha wrote:
$$x=1$$. So, set $$M$$ consists of only one element.

The range of a single element set is 0.

I didn't understand from here on. What's the logic? I know that range is the output, but how do we determine it?[/quote]

The range is the difference between the largest and smallest elements of a set. The range of a single element set is therefore 0 because in a single element set the largest and smallest elements are the same.
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Re: M18-35 &nbs [#permalink] 11 Jul 2018, 11:14
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# M18-35

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