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Bunuel
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Firstly, I started solving this way:
x^2 − 2x = 3
x (x - 2) = 3
Either x = 3,
or
x - 2 = 3
=> x = 5

While noticing that putting 5 in the original equation (25 - 10) doesn't equal to 3, l realized l could put all values in one side x^2 − 2x − 3 = 0 to derive at x = 3 or -1, and then answer the question.

l was wondering what's wrong with the approach l started with. Could someone clarify please?

Thank you.
Regards.
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ranaazad
Firstly, I started solving this way:
x^2 − 2x = 3
x (x - 2) = 3
Either x = 3,
or
x - 2 = 3
=> x = 5

While noticing that putting 5 in the original equation (25 - 10) doesn't equal to 3, l realized l could put all values in one side x^2 − 2x − 3 = 0 to derive at x = 3 or -1, and then answer the question.

l was wondering what's wrong with the approach l started with. Could someone clarify, please?

Thank you.
Regards.


Hi Ranaazad,
This way is a bit faulty.

When you say x (x - 2) = 3 then x = 3/(x-2) or (x - 2) = 3/x.

What you're ignoring is that you have integer 3 at the right-hand side, not ZERO.
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x^2 − 2x = 3
=>x^2 − 2x - 3 = 0
=>x^2 − 3x + x - 3 = 0
=>x(x − 3) + 1(x - 3) = 0
=>(x − 3)(x + 1) = 0
therefore
x=-1 or 3
Check with options keeping x=-1 or 3

A. x^2 − 6x + 9 => will yield 0 when x=3
B. x^2 − 4x + 3 => will yield 0 when x=3
C. x^2 − x − 2 => will yield 0 when x=-1
D. x^2 − 7x + 6 => will not yield 0 when x=-1 or 3 Hence answer D
E. x^2 − 9 => will yield 0 when x=3
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