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If (x – y) = and (x + y) = , what is the value of x2 – y2
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17 Jul 2018, 11:50
17 Jul 2018, 11:50
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Expert Reply
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Question Stats:
83% (00:29) correct
16% (00:57) wrong based on 91 sessions
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If \((x - y) =\sqrt{12}\) and \((x + y) = \sqrt{3}\), what is the value of \(x^2 - y^2\)?
(A) 3
(B) 6
(C) 9
(D) 36
(E) It cannot be determined from the information given.
(A) 3
(B) 6
(C) 9
(D) 36
(E) It cannot be determined from the information given.
Re: If (x – y) = and (x + y) = , what is the value of x2 – y2? (
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18 Jul 2018, 07:16
18 Jul 2018, 07:16
2
Expert Reply
Explanation
The factored form of the Difference of Squares (one of the “special products” you need to memorize for the exam) is comprised of the terms given in this problem:
\(x^2 - y^2 = (x + y)(x - y)\)
Substitute the values \(\sqrt{12}\) and \(\sqrt{3}\) in place of (x – y) and (x + y), respectively:
x^2 - y^2 = \(\sqrt{12} \times \sqrt{3}\)
Combine 12 and 3 under the same root sign and solve:
\(x^2 - y^2 =\sqrt{12} \times \sqrt{3}\)
\(x^2 - y^2 =\sqrt{36}\)
\(x^2 - y^2 = 6\)
The factored form of the Difference of Squares (one of the “special products” you need to memorize for the exam) is comprised of the terms given in this problem:
\(x^2 - y^2 = (x + y)(x - y)\)
Substitute the values \(\sqrt{12}\) and \(\sqrt{3}\) in place of (x – y) and (x + y), respectively:
x^2 - y^2 = \(\sqrt{12} \times \sqrt{3}\)
Combine 12 and 3 under the same root sign and solve:
\(x^2 - y^2 =\sqrt{12} \times \sqrt{3}\)
\(x^2 - y^2 =\sqrt{36}\)
\(x^2 - y^2 = 6\)
If (x y) = and (x + y) = , what is the value of x2 y2
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15 Jun 2023, 03:14
15 Jun 2023, 03:14
1
\(x^2 - y^2 = (x-y)(x+y)\)
\(x^2 - y^2 = \sqrt{12} * \sqrt{3}\)
\(x^2 - y^2 = \sqrt{12 * 3} = \sqrt{4*3*3} = 3*\sqrt{4} = 3*2 = 6\)
Answer is Choice B
\(x^2 - y^2 = \sqrt{12} * \sqrt{3}\)
\(x^2 - y^2 = \sqrt{12 * 3} = \sqrt{4*3*3} = 3*\sqrt{4} = 3*2 = 6\)
Answer is Choice B
