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Question
"I got on a test question that:
3x^2 - 2x + 7 = 0, then ( x - 1/3 )^2 = ???

The correct is -20/9.  I don't see how this is possible since you get a negative number under the radical sign (-80) when using the quadratic formula.
Could you explain this? Thanks...."

Replace the ??? with a variable, say y.  This gives (x - 1/3)^2 = y.

Squaring the left side of this equation gives x^2 - 2x/3 + 1/9 = y.
Multiplying the entire equation by 3 gives 3x^2 - 2x + 1/3 = 3y.
Subtracting 1/3 from both sides gives 3x^2 - 2x = 3y - 1/3.

The first equation is 3x^2 - 2x + 7 = 0.
Subtracting 7 from both sides gives 3x^2 - 2x = -7.

Since both equations have 3x^2 - 2x on the left,
the right sides can be set equal to each other.

This says that 3y - 1/3 = -7.  Note that -7 = -21/3.
Add 1/3 to both sides gives 3y = -20/3.
Dividing the both sides by 3 gives y = -20/9.

Putting this into (x - 1/3)^2 = y and checking it out gives (x - 1/3)^2 = -20/9.
Squaring the left side gives x^2 - 2x/3 + 1/9 = -20/9.
Adding 20/9 to both sides gives x^2 - 2x/3 + -21/9 = 0.

Note that 21/9 reduces to 7/3, so we have x^2 - 2x/3 + 7/3 = 0.
Multiplying this equation by 3 gives 3x^2 - 2x + 7 = 0.

The 2nd equation was transformed into the original, so it checks out.

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