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If b=a+4, then for which of the following . . .

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If b=a+4, then for which of the following . . .

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If b=a+4, then 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

The OA i the option C.

Is there a fast way to get the correct answer? Should I try option by option? Experts, can you help me? Thanks in advanced.

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Hello M7MBA.

Let's take a look at your question.

If b=a+4 then $$\left(x-a\right)^2+\left(x-b\right)^2=\left(x-a\right)^2+\left(\left(x-a\right)-4\right)^2=\left(x-a\right)^2+\left(x-a\right)^2-8\left(x-a\right)+16$$ $$=2\left(x-a\right)^2-8\left(x-a\right)+16=2\left(\left(x-a\right)^2-4\left(x-a\right)-8\right).$$ Now,

If x=a-1 then we will get $$2\left(1+4-8\right)=-6.$$ If x=a then $$2\left(-8\right)=-16.$$ If x=a+2 the $$2\left(4-8-8\right)=-24.$$ If x=a+3 the $$2\left(9-12-8\right)=-22.$$ And finally if x=a+5 then $$2\left(25-20-8\right)=-6.$$ So, the smallest value is -24. Therefore, the correct answer is C=a+2.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.

_________________
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M7MBA wrote:
If b=a+4, then 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

The OA i the option C.

Is there a fast way to get the correct answer? Should I try option by option? Experts, can you help me? Thanks in advanced.
Algebraically (a+b)^2+(a-b)^2=2(a^2+b^2)
Or (a^2+b^2) = ½[(a+b)^2+(a-b)^2]
Using this formula over the given expression we get
(x-a)^2+(x-b)^2=(1/2)[(2x-a-b)^2+(b-a)^2]
Put b=a+4
=1/2[(2x-a-a-4)^2+(a+4-a)^2]
=1/2[(2x-2a-4)^2+4^2]
=(1/2)* 4[(x-a-2)^2 + 4]=2(x-a-2)^2 + 8
for this to be minimum first expression 2[x-a-2]^2 should be equal to zero
which makes x=a+2
hence option C

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