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

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

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New post Updated on: 19 Nov 2012, 02:54
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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, 02:49.
Last edited by Bunuel on 19 Nov 2012, 02:54, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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New post 19 Nov 2012, 03:00
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.

Answer: A.

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

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New post 19 Nov 2012, 03:32
thank you, very helpful
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Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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New post 24 May 2018, 13:24
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.

Answer: A.

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

\(x = 9\)
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If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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New post 24 May 2018, 21:10
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.

Answer: A.

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!
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Re: If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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New post 25 May 2018, 12:47
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.

Answer: A.

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 Bunuel`s approach, is there another way to solve this problem ?

have a good weekend :-)
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If set M consists of the root(s) of equation 2-x^2 = (x-2)^2  [#permalink]

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New post 25 May 2018, 17:43
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.

Answer: A.

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 Bunuel`s 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!
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If set M consists of the root(s) of equation 2-x^2 = (x-2)^2 &nbs [#permalink] 25 May 2018, 17:43
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