If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =

mejia401 , your solution would have been fine had there not been the last option of "none of the above". You were lucky to get the answer but there might be instances where the actual answer would be "none of the above" and in this case, you will invariably mark the incorrect answer.

\(y = (x – 5)^2 + (x + 1)^2 – 6 = 2x^2-8x+20 = 2x^2-8x+8 + 12\)(try to complete a "\((a \pm b)^2\)")

Thus, \(y = 2(x^2-4x+4) + 12 = 2(x-2)^2 + 12\).

Now, any square of the form \((a-b)^2\) will be MINIMUM, when a=b and the minimum value = 0 ( as for any square , \(x^2 \geq 0\), thus minimum value =0 ).

Based on the discussion above,

minimum value of \(y = 2(x-2)^2 + 12\) , will be when x = 2 and at that value of x, y = 12.
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800score Official Solution:

Let us transform the formula:
y = (x – 5)² + (x +1)² – 6 =
x² – 10x + 25 + x² + 2x + 1 – 6 =
2x² – 8x + 20 = 2 × (x² – 4x + 10) =
2 × ((x² – 4x + 4) + 6) =
2 × ((x – 2)² + 6)

Any square is greater or equal 0. Therefore the formula possess the least value when (x – 2)² = 0.
x – 2 = 0
x = 2

The correct answer is choice (D).
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For a lack of better methods, I lined up each answer choices and input them as a function of \(x\)
A. -2 36<
B. -1 36
C. 0 25
D. 2 18
E. None of the above

IMO D.

I tested 3 and -3.
If x = 3, y = 14
If x = -3, y = 62

This tells me x is probably closer to 3. So I start with 2 to get 12 for y. Then I test x = 1, and get y = 14.

Therefore I know 2 is the correct answer because any number smaller than 2 makes y larger and any number larger than 2 also does the same.

Ans D

I use differentiation.
y = (x – 5)^2 + (x + 1)^2 – 6
y'=0=2(x-5)+2(x+1)
x=2

\((x-5)^2 + (x+1)^2 - 6 = x^2 - 10x + 25 + x^2 + 2x + 1 - 6 = 2x^2 - 8x + 20 = y.\)

\(\frac{dy}{dx} = 4x - 8 = 0 --> x = 2\). Ans (D).
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Expanding and then differentiating is probably the quickest. Just have to do a quick check to make sure that you got the minimum, as required by the question, and not the maximum.

Option:D
Time Taken:1:34
here you go
y=(x-5)^2+(x+1)^2-6[expand using formula(a-b)^2 and (a+b)^2]
y=x^2-10x+25+x^2+2x+1-6
y=2x^2-8x+20
Divide by
y=x^2-4x+10
Since we need to find least so putting negative values wont make sense keeping in powers and negative sign with co-efficient
start with "0"
y=10
then "2"
y=4-8+10
y=6

Let’s simplify the equation first:

y = (x – 5)^2 + (x + 1)^2 – 6

y = x^2 - 10x + 25 + x^2 + 2x + 1 - 6

y = 2x^2 - 8x + 20

We see that the equation is an upward parabola, so the minimum occurs at the vertex of the parabola. That is, we need to find the x-value of the vertex, for which we can use the formula x = -b/(2a):

x = -(-8)/(2(2)) = 8/4 = 2

Answer: D
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dy/dx = 2(x-5) + 2(x+1) = 0
2x -4 =0
x = 2

IMO D

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Dear Bunuel,

I'm stuck at that part where you split 10 into 4 + 6 (the part that I highlighted). How do we know how to split a number to get the form of (a+b)^2?

Thank you!

Dear Bunuel,

Never mind - after a night of sleep I figured it out this morning. Thanks!

y = (x – 5)^2 + (x + 1)^2 – 6 = 2x^2 - 8x + 20

y = ax^2 + bx + c
In this case: a = 2>0 => The parabola upward
=> y is least at the vertex of a parabola
=> y = -b/2a = 8/(2*2) = 2
=> OA is D

Hope this helps

i don't know if i'm right but i found the vertex of a parabola for the equation ax^2+bx+c using X=-b/2a
X=8/4=2 this is the x cordinate of the vertex and should be the minimum

very fast

A lot of these solutions seemed a little too complex for me and the following 2 approaches seemed doable for people like me -
Approach 1
Plug in answers
You don't even have to calculate final results for all. For example
A) 49+1-6
B) 36-6=30 (Thus, A is clearly out)
C) 25+1-6=20 (Thus B is clearly out)
D) 9+9+6=12 (Winner)
Answer : D

Approach 2 - Logic
Goal : Minimize y
On expansion of the solution, I stopped at -
2(x^2 - 4x +10)
Now notice, logically, in order to minimise y -
1. Best scenario is if you are able to eliminate x all together and are left with 2*10 = 20. Thus if x=0, y=20
2. Analyse that any negative value of x will make -4x and x^2 a positive number which will actually increase the value of y which is the opposite of what we want. Eg: If x=-1 => 1+4+10 = larger value of y
3. Now from 1, we can take stock of the fact that the next approach to further reduce the value of 10 would be if somehow we can leverage -4x and subtract from 10. This is possible for a positive number greater than 0. Scan the options, and you will find only one positive value greater than 0.
Answer : D

Hope this helps.

We can do it by using differentiation as well.
we know y is min when its differentiation is zero.
dy/dx=0 which means 2(x-5)(1)+2(x+1)(1)+0 = 0

4x=8
and therefore x=2

Similar question: https://gmatclub.com/forum/find-the-lea ... 46220.html

If y=(x–5)2+(x+1)2–6y=(x–5)2+(x+1)2–6, then y is least when x =

A. -2
B. -1
C. 0
D. 2
E. None of the above

Expanding we get :X^2 + 25 - 10X + X^2 +2X + 1 - 6
=2X^2 - 8X + 20
=2(X^2 - 4X + 10)
For a quadratic equation of the form AX^2 + BX + C Where A>0, the minimum value of the function occurs at X =\( -B/2A\)
This means for the given function minimum value of Y occurs at X = \(-(-4)/2(1)\)= \(4/2\) =2
D

Hi Bunuel - can you confirm why are we transforming the equation to (x-2)^2? to get the minimum values?