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For all integers a and b, where a ≠ b

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Carcass
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For all integers a and b, where a ≠ b [#permalink] New post   10 Mar 2019, 06:34
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For all integers a and b, where \(a ≠ b\), \(a ★ b = \mid ( \frac{a^2 - b^2}{(a - b)} \mid\)

What is the value of 4★2 ?

A. 2

B. 4

C. 6

D. 8

E. 10
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C
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MagooshStudentHelp
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Re: For all integers a and b, where a ≠ b [#permalink] New post   17 Mar 2019, 17:57
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All we do here is plug a = 4 and b = 2 into the given expression. So:

\(\frac{4^2 - 2^2}{4 - 2}\)

Simplify:

\(\frac{16 - 4}{2}\)

\(\frac{12}{2} = 6\)
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Re: For all integers a and b, where a ≠ b [#permalink] New post   14 Dec 2021, 00:57
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Re: For all integers a and b, where a ≠ b [#permalink] New post   14 Dec 2021, 06:23
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Carcass wrote:
For all integers a and b, where \(a ≠ b\), \(a ★ b = \mid ( \frac{a^2 - b^2}{(a - b)} \mid\)

What is the value of 4★2 ?

A. 2
B. 4
C. 6
D. 8
E. 10


One approach is to recognize that we can SIMPLIFY the fraction \(\frac{a^2 - b^2}{(a - b)}\) by first factoring the numerator.

We get: \(|\frac{a^2 - b^2}{(a - b)}| = |\frac{(a+b)(a-b)}{(a - b)}|=|a+b|\)

So, a ★ b = |a + b|
This means: 4 ★ 2 = |4 + 2| = |6| = 6

Cheers,
Brent
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Re: For all integers a and b, where a b [#permalink] New post   24 Aug 2022, 09:37
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Given that \(a ★ b = |\frac{(a^2 - b^2)}{(a - b)}|\) and we need to find the value of \(4 ★ 2\)

To find \(4 ★ 2\) we need to compare what is before and after ★ in \(4 ★ 2\) and a ★ b

=> We need to substitute a with 4 and b with 2 in a ★ b to get the value of \(4 ★ 2\)

=> \(4 ★ 2 = |\frac{(4^2 - 2^2)}{(4 - 2)}|\) = \(|\frac{16 - 4}{2}|\) = \(|\frac{12}{2}|\) = \(|6|\) = 6
( Watch this video to learn about the Basics of Absolute Value )

So, Answer will be C
Hope it helps!

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Re: For all integers a and b, where a b [#permalink]
24 Aug 2022, 09:37
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