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Find all pairs of real numbers (a, b) such that (x-a)^2 + (2x-b)^2 = (x-3)^2 + (2x)^2 for all x.

 Feb 15, 2022
 #1
avatar+117175 
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\((x-a)^2 + (2x-b)^2 = (x-3)^2 + (2x)^2\)

 

I'd start by expanding it out.  Have you done that?  Show us.

 Feb 15, 2022
 #2
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I get \(x^2+a^2-2ax + 4x^2+b^2-4bx= x^2 + 9 - 6x + 4x^2\)

Guest Feb 15, 2022
 #3
avatar+117175 
+1

Nice work  guest !

 

\(x^2+a^2-2ax + 4x^2+b^2-4bx= x^2 + 9 - 6x + 4x^2\)

 

Now simply the left side and simplify the right side independently of each other.

 

\(x^2+a^2-2ax + 4x^2+b^2-4bx= x^2 + 9 - 6x + 4x^2\\ 5x^2+(-2a-4b)x+(a^2+b^2)=5x^2-6x+9\)

 

Now equate coefficients

\(5=5\\~\\ -2a-4b=-6\\ a+2b=3\\ a=3-2b \\~\\ (3-2b)^2+b^2=9\\ 9+4b^2-12b+b^2=9\\ 5b^2-12b=0\\ b(5b-12)=0\\ b=0\quad or \quad b=12/5 = 2.4\\~\\ if\; b=0, \\\quad a=3-0=3\\ if\;b=2.4, \\\quad a=3-2*2.4=3-4.8=-1.8\\ \text{so I have }(0,3)\quad and \quad (2.4,-1.8) \)

 

You need to check these by plugging them into the ORIGINAL equation.

Melody  Feb 16, 2022

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