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

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Math Expert
Joined: 02 Sep 2009
Posts: 51218
If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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29 Jul 2015, 14:49
1
4
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:15) correct 25% (01:03) wrong based on 301 sessions

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If y = (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

Kudos for a correct solution.

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

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29 Jul 2015, 15:16
Bunuel wrote:
If y = (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

Kudos for a correct solution.

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.
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If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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29 Jul 2015, 17:16
1
Bunuel wrote:
If y = (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

Kudos for a correct solution.

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
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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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29 Jul 2015, 18:24
2
2
mejia401 wrote:
Bunuel wrote:
If y = (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

Kudos for a correct solution.

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.

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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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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01 Aug 2015, 06:10
1
I use differentiation.
y = (x – 5)^2 + (x + 1)^2 – 6
y'=0=2(x-5)+2(x+1)
x=2
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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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01 Aug 2015, 09:07
3
$$(x-5)^2 + (x+1)^2 - 6 = x^2 - 10x + 25 + x^2 + 2x + 1 - 6 = 2x^2 - 8x + 20 = y.$$

$$\frac{dy}{dx} = 4x - 8 = 0 --> x = 2$$. Ans (D).
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Posts: 51218
Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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17 Aug 2015, 08:57
Bunuel wrote:
If y = (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

Kudos for a correct solution.

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

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16 Sep 2015, 12:07
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.
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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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20 Jul 2017, 01:05
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
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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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25 Jul 2017, 10:47
Bunuel wrote:
If y = (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

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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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x =  [#permalink]

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02 Dec 2018, 03:05
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Re: If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when x = &nbs [#permalink] 02 Dec 2018, 03:05
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