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

dave13Bunuel'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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