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I think this is a high-quality question and I agree with explanation. Assume a = 0
(x)2+(x−4)2

-1 --- 10
0 --- 16
2 --- 8
3 --- 10
5 ---- 26

So the answer is option C
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I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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We are give \((x-a)^2-(x-b)^2\), and since both are squared values, the least possible value for the whole expression can be 0.
Now doing \((x-a)^2-(x-b)^2=0\). Now \((x-a)^2=-(x-b)^2\).
Ignore this -ve sign(as LHS is square number and RHS can't be -ve), and take square root on both sides.
There can be 2 cases:-
1) \(x-a=x-b=0\) (this does not give us any value of x, thus ignore this case.)
2) \(x-a=b-x\) , we get \(2x=b+a\), replace b with a+4 as given, we get \(x=a+2\).
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Hi, Ricky..

How do you get in the last step that x-a=b-x?
Why you only consider the negative root for only 1 side of the equation.

Regards,

Juan C. Avellan
rickyric395
We are give \((x-a)^2-(x-b)^2\), and since both are squared values, the least possible value for the whole expression can be 0.
Now doing \((x-a)^2-(x-b)^2=0\). Now \((x-a)^2=-(x-b)^2\).
Ignore this -ve sign(as LHS is square number and RHS can't be -ve), and take square root on both sides.
There can be 2 cases:-
1) \(x-a=x-b=0\) (this does not give us any value of x, thus ignore this case.)
2) \(x-a=b-x\) , we get \(2x=b+a\), replace b with a+4 as given, we get \(x=a+2\).
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avellanjc
Hi, Ricky..

How do you get in the last step that x-a=b-x?
Why you only consider the negative root for only 1 side of the equation.

Regards,

Juan C. Avellan
rickyric395
We are give \((x-a)^2-(x-b)^2\), and since both are squared values, the least possible value for the whole expression can be 0.
Now doing \((x-a)^2-(x-b)^2=0\). Now \((x-a)^2=-(x-b)^2\).
Ignore this -ve sign(as LHS is square number and RHS can't be -ve), and take square root on both sides.
There can be 2 cases:-
1) \(x-a=x-b=0\) (this does not give us any value of x, thus ignore this case.)
2) \(x-a=b-x\) , we get \(2x=b+a\), replace b with a+4 as given, we get \(x=a+2\).

That solution is not precise, so I'd ignore it.
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Although I did substitute the options and solved this during the test, while reviewing, I saw an algebra approach, so if someone is looking for one:

\((x - a)^2 + (x - b)^2\)
= \((x - a)^2 + (x - a - 4)^2\)
= \((x - a)^2 + ((x - a) - 4)^2\)
Take x - a = k
= \(k^2 + (k - 4)^2\)
= \(k^2 + k^2 + 16 - 8k\)
= \(2k^2 - 8k + 16\)
= \(2(k^2 - 4k + 8)\)
= \( 2(k^2 - 4k + 4 + 4) \)
= \( 2(k - 2)^2 + 8 \)
Since a square of a term is always greater than or equal to 0, the minimum value of this expression is 8, and it occurs when
\( 2(k - 2)^2 = 0 \), which means k = 2, which means, x - a = 2, and x = a + 2.

Hope it helps.
Bunuel
If \(b = a + 4\), for which of the following values of \(x\) is the expression \((x - a)^2 + (x - b)^2\) the smallest?

A. \(a-1\)
B. \(a\)
C. \(a + 2\)
D. \(a + 3\)
E. \(a + 5\)
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