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Re: If (x-1)^2=(x-2)^2, then x= [#permalink]
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(x-1)^2 = (x-2)^2

We know that quadratic curves are symmetric. so what number ends up same magnitude when you subtract 1 and 2 from it?

1.5 because 1.5-1 = .5 and 1.5-2 = -.5. (squaring cancels the negative 1)
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Re: If (x-1)^2=(x-2)^2, then x= [#permalink]
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if \((x-1)^ = (x-2)^2\) then,

expanding both sides of the equation, we get

\(x^2 + 1^2 - 2\times 1\times x = x^2 + 2^2 - 2\times 2 \times x\)

\(x^2 + 1 - 2x = x^2 + 4 - 4x\)

cancelling like terms on both sides and rearranging,

\(2x - 3 = 0\)

\(2x = 3\)

\(x = \frac{3}{2}\)
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Re: If (x-1)^2=(x-2)^2, then x= [#permalink]
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Hi

There is a typo in option D. It should be 3/2.

Thanks

Carcass wrote:
If \((x-1)^2=(x-2)^2\), then \(x=\)

A. -5/8

B. 5/3

C. 4/3

D. 2/3

E. 5/2
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Re: If (x-1)^2=(x-2)^2, then x= [#permalink]
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the book had the typo

thanks sir fixed
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Re: If (x-1)^2=(x-2)^2, then x= [#permalink]
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