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# (x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =

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(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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11 Jun 2018, 22:38
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15% (low)

Question Stats:

94% (01:20) correct 6% (02:11) wrong based on 66 sessions

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$$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =$$

(A) $$5x^2$$

(B) $$5x^2 + 10$$

(C) $$x^2 + 10$$

(D) $$5x^2 + 6x + 10$$

(E) $$5x^2 - 6x + 10$$

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Re: (x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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11 Jun 2018, 22:47
Bunuel wrote:
$$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =$$

(A) $$5x^2$$

(B) $$5x^2 + 10$$

(C) $$x^2 + 10$$

(D) $$5x^2 + 6x + 10$$

(E) $$5x^2 - 6x + 10$$

Instead of jumping straight into the calculation, we'll first look at the answers.
This is an Alternative approach.

All of our answers are very similar - they are mostly quadratic equations of the form ax^2 + bx + c.
Let's look at each coefficient separately:
According to the answers a can be 1 or 5.
Looking at our equation, we see that we have 5 expressions, each of which gives one copy of x^2.
So a = 5.
(C) is eliminated.
b can be either 0, 6 or -6.
Looking at our expressions, we can see that our expression is symmetrical: (x - 2)^2 and (x+2)^2 cancel out as do (x - 1)^2 and (x + 1)^2.
So b must be 0. (D), (E) are eliminated.
Finally, c is 0 or 10. As it is the sum of positive numbers (2^2, 1^2, 1^2, 2^2), it can't be 0.
(A) is eliminated.

Note that it is also perfectly possible to just straight up calculate the value of the expression.
Using the answers to guide your calculation can help avoid making silly mistakes.
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Re: (x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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11 Jun 2018, 22:48

Solution

To find:
• The value of the expression $$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =$$

Approach and Working:
We know, $$(a + b)^2 + (a – b)^2 = 2 (a^2 + b^2)$$

Using this we can rewrite the given expression as
• $$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2$$
= $$(x + 2)^2 + (x - 2)^2 + (x + 1)^2 + (x - 1)^2 + x^2$$
= $$2 (x^2 + 4) + 2 (x^2 + 1) + x^2$$
= $$2x^2 + 8 + 2x^2 + 2 + x^2$$
= $$5x^2 + 10$$

Hence, the correct answer is option B.

Note: One can also get the answer using value substitution
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Re: (x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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12 Jun 2018, 00:55
Bunuel wrote:
$$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =$$

(A) $$5x^2$$

(B) $$5x^2 + 10$$

(C) $$x^2 + 10$$

(D) $$5x^2 + 6x + 10$$

(E) $$5x^2 - 6x + 10$$

(a - b)^2 = a^2 - 2ab + b^2 | (a + b)^2 = a^2 + 2ab + b^2

If we add them we will get (a - b)^2 + (a + b)^2 = 2a^2 + 2b^2

(x - 2)^2 + (x + 2)^2 = 2x^2 + 2(2^2) = 2x^2 + 8
(x - 1)^2 + (x + 1)^2 = 2x^2 + 2(1^2) = 2x^2 + 2

Therefore, $$(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 = 2x^2 + 8 + x^2 + 2x^2 + 2 = 5x^2 + 10$$(Option B)
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(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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12 Jun 2018, 01:05
By alternative approach we can first check the coefficient of X as that will help to eliminate more than one option. From the equation we see that its -4x,-2x,4x,2x i.e overall coefficient of X is 0 thus that eliminates D and E. Similarly we see that the coefficient of x^2 is 5 thus we eliminate C and finally since in(a+-b)^2 b^2 will always be positive except when b=0 thus total b^2 can’t be 0 thus that eliminates A.. thus B is the answer

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(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =  [#permalink]

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12 Jun 2018, 05:08
$$(1 - 2)^2 + (1 - 1)^2 + 1^2 + (1 + 1)^2 + (1 + 2)^2 =15$$

(A) $$5x^2$$
(B) $$5*1^2 + 10=15$$
(C) $$x^2 + 10$$
(D) $$5x^2 + 6x + 10$$
(E) $$5x^2 - 6x + 10$$

Just plug in 1 instead of X, then only option B holds true. Don't mess with algebra until it is necessary.
(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =   [#permalink] 12 Jun 2018, 05:08
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