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(x - 2)2 + (x - 1)2 + x2 + (x + 1)2 + (x + 2)2 =
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19 May 2018, 15:22
19 May 2018, 15:22
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Question Stats:
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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) \(5x^2 - 6x + 10\)
(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 + x2 + (x + 1)2 + (x + 2)2 =
[#permalink]
30 Dec 2018, 11:43
30 Dec 2018, 11:43
sandy 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) \(5x^2\)
(B) \(5x^2 + 10\)
(C) \(x^2 + 10\)
(D) \(5x^2 + 6x + 10\)
(E) \(5x^2 - 6x + 10\)
One option is to use FOIL to expand each expression, and then simplify everything.
I'll let someone else demonstrate that solution.
Another approach is to recognize that we're looking for an EQUIVALENT EXPRESSION
So, if we evaluate \((x - 2)^2 + (x - 1)^2 + x^2 + (x + 1)^2 + (x + 2)^2\) for a certain value of x, then its EQUIVALENT EXPRESSION must also evaluate to the same number for that same x-value.
For example, if x = 1, then: (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² = (1 - 2)² + (1 - 1)² + 1² + (1 + 1)² + (1 + 2)²
= (-1)² + (0)² + 1 + (2)² + (3)²
= 1 + 0 + 1 + 4 + 9
= 15
So, when x = 1, the original expression evaluates to be 15
Now we'll check the answer choices to see which one(s) evaluate to 15, when x = 1
We get:
(A) 5x² = 5(1²)= 5 We need the expression to evaluate to be 15. ELIMINATE
(B) 5x² + 10 = 5(1²) + 10 = 15 PERFECT! Keep.
(C) x² + 10 = (1²) + 10 = 11 We need the expression to evaluate to be 15. ELIMINATE
(D) 5x² + 6x + 10 = 5(1²) + 6(1) + 10 = 21 We need the expression to evaluate to be 15. ELIMINATE
(E) 5x² - 6x + 10 = 5(1²) - 6(1) + 10 = 9 We need the expression to evaluate to be 15. ELIMINATE
Answer: B
Cheers,
Brent
gmatclubot
Re: (x - 2)2 + (x - 1)2 + x2 + (x + 1)2 + (x + 2)2 = [#permalink]
30 Dec 2018, 11:43
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