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

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Joined: 02 Sep 2009
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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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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.

(B) is our answer.

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.

Answer: 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.
Answer (B)
(x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =   [#permalink] 12 Jun 2018, 05:08
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# (x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2 =

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