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

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).

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

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

IMO D

Posted from my mobile device

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?