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# If set M consists of the root(s) of equation 2-x^2 = (x-2)^2

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Intern
Joined: 01 Apr 2012
Posts: 2
If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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Updated on: 19 Nov 2012, 01:54
2
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Difficulty:

45% (medium)

Question Stats:

65% (01:40) correct 35% (01:45) wrong based on 296 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

M18-36

I am having trouble with the below question

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

2−x^2=(x−2)^2;

2−x^2=x^2−4x+4; (how do I get from here to the next row?)

x^2−2x+1=0;

(x−1)^2=0;

Thanks,
Jennifer

Originally posted by J270 on 19 Nov 2012, 01:49.
Last edited by Bunuel on 19 Nov 2012, 01:54, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 50729
Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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19 Nov 2012, 02:00
2
1
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

$$2-x^2 = (x-2)^2$$ --> $$2-x^2=x^2-4x+4$$ --> add (x^2-2) to both sides: $$2-x^2+(x^2-2)=x^2-4x+4+(x^2-2)=0$$ --> $$0=2x^2-4x+2$$ --> reduce by 2: $$0=x^2-2x+1$$ --> $$(x-1)^2=0$$ --> $$x=1$$. So, set M consists of only one element.

The range of a single element set is 0.

Hope it's clear.
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Joined: 01 Apr 2012
Posts: 2
Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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19 Nov 2012, 02:32
VP
Joined: 09 Mar 2016
Posts: 1111
Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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24 May 2018, 12:24
Bunuel wrote:
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

$$2-x^2 = (x-2)^2$$ --> $$2-x^2=x^2-4x+4$$ --> add (x^2-2) to both sides: $$2-x^2+(x^2-2)=x^2-4x+4+(x^2-2)=0$$ --> $$0=2x^2-4x+2$$ --> reduce by 2: $$0=x^2-2x+1$$ --> $$(x-1)^2=0$$ --> $$x=1$$. So, set M consists of only one element.

The range of a single element set is 0.

Hope it's clear.

Bunuel why is my approach wrong ? i followd all algebra rules

$$(x-2)^2$$ for this i used formula $$(a-b)^2 = (a-b) (a+b)$$

hence $$2-x^2 = (x-2)^2 --> 2-x^2 = (x-2)(x+2) = x^2+2x-2x-4 = x^2 - 4$$

$$2-x^2 =x^2 - 4$$

$$2x^2 = 6$$

$$x^2= 3$$ square both parts

$$x = 9$$
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3311
Location: India
GPA: 3.12
If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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24 May 2018, 20:10
1
dave13 wrote:
Bunuel wrote:
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

$$2-x^2 = (x-2)^2$$ --> $$2-x^2=x^2-4x+4$$ --> add (x^2-2) to both sides: $$2-x^2+(x^2-2)=x^2-4x+4+(x^2-2)=0$$ --> $$0=2x^2-4x+2$$ --> reduce by 2: $$0=x^2-2x+1$$ --> $$(x-1)^2=0$$ --> $$x=1$$. So, set M consists of only one element.

The range of a single element set is 0.

Hope it's clear.

Bunuel why is my approach wrong ? i followd all algebra rules

$$(x-2)^2$$ for this i used formula $$(a-b)^2 = (a-b) (a+b)$$

hence $$2-x^2 = (x-2)^2 --> 2-x^2 = (x-2)(x+2) = x^2+2x-2x-4 = x^2 - 4$$

$$2-x^2 =x^2 - 4$$

$$2x^2 = 6$$

$$x^2= 3$$ square both parts

$$x = 9$$

Hey dave13

Unfortunately, the formula you have used is wrong: (a-b)^2 = (a-b)(a+b)

The correct formulae are
1. $$(a-b)^2 = a^2 - 2ab + b^2$$
2. $$a^2 - b^2 = (a-b)(a+b)$$

Hope this helps you!
_________________

You've got what it takes, but it will take everything you've got

VP
Joined: 09 Mar 2016
Posts: 1111
Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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25 May 2018, 11:47
Bunuel wrote:
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

$$2-x^2 = (x-2)^2$$ --> $$2-x^2=x^2-4x+4$$ --> add (x^2-2) to both sides: $$2-x^2+(x^2-2)=x^2-4x+4+(x^2-2)=0$$ --> $$0=2x^2-4x+2$$ --> reduce by 2: $$0=x^2-2x+1$$ --> $$(x-1)^2=0$$ --> $$x=1$$. So, set M consists of only one element.

The range of a single element set is 0.

Hope it's clear.

pushpitkc many thanks for explanation one question how from this $$0=x^2-2x+1$$ we got this $$(x-1)^2=0$$ ?

in addittion to Bunuels approach, is there another way to solve this problem ?

have a good weekend
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3311
Location: India
GPA: 3.12
If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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25 May 2018, 16:43
dave13 wrote:
Bunuel wrote:
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

$$2-x^2 = (x-2)^2$$ --> $$2-x^2=x^2-4x+4$$ --> add (x^2-2) to both sides: $$2-x^2+(x^2-2)=x^2-4x+4+(x^2-2)=0$$ --> $$0=2x^2-4x+2$$ --> reduce by 2: $$0=x^2-2x+1$$ --> $$(x-1)^2=0$$ --> $$x=1$$. So, set M consists of only one element.

The range of a single element set is 0.

Hope it's clear.

pushpitkc many thanks for explanation one question how from this $$0=x^2-2x+1$$ we got this $$(x-1)^2=0$$ ?

in addittion to Bunuels approach, is there another way to solve this problem ?

have a good weekend

Hey dave13

Bunuel's approach is the easiest method to solve this problem. All we do is expand the equation and solve the quadratic equation to find the elements of the set. Once we find that, we can
calculate range, which is the difference between the maximum and minimum value of the set

Now, coming back to your question, two formulae are as follows
$$a^2 - 2ab + b^2 = (a - b)^2$$ | $$a^2 + 2ab + b^2 = (a + b)^2$$

In the equation $$x^2-2x+1 = 0$$

a = x, b = 1, and 2ab = -2x | Now, $$x^2 - 2(x)(1) + 1^2 = (x - 1)^2$$

Hope this helps you!
_________________

You've got what it takes, but it will take everything you've got

If set M consists of the root(s) of equation 2-x^2 = (x-2)^2 &nbs [#permalink] 25 May 2018, 16:43
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