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\((\sqrt{245}-\sqrt{75})^2=\)

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by VJesus12 » Mon May 11, 2020 6:40 am
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\((\sqrt{245}-\sqrt{75})^2=\)


A. \(170−5\sqrt8\)

B. \(170−70\sqrt{15}\)

C. \(320−70\sqrt{15}\)

D. \(318−35\sqrt{15}\)

E. \(170\)

[spoiler]OA=C[/spoiler]

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Re: \((\sqrt{245}-\sqrt{75})^2=\)

by Brent@GMATPrepNow » Mon May 11, 2020 9:16 am
VJesus12 wrote: ↑
Mon May 11, 2020 6:40 am
 \((\sqrt{245}-\sqrt{75})^2=\)

A. \(170−5\sqrt8\)
B. \(170−70\sqrt{15}\)
C. \(320−70\sqrt{15}\)
D. \(318−35\sqrt{15}\)
E. \(170\)
√245 = (√49)(√5) = 7√5
√75 = (√25)(√3) = 5√3

So, (√245 - √75 )² = (7√5 - 5√3)²
= (7√5 - 5√3)(7√5 - 5√3)
= 49√25 - 35√15 - 35√15 + 25√9
= (49)(5) - 35√15 - 35√15 + (25)(3)
= 245 - 35√15 - 35√15 + 75
= 320 - 70√15

Answer: C
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Re: \((\sqrt{245}-\sqrt{75})^2=\)

by Scott@TargetTestPrep » Wed May 13, 2020 8:15 am
VJesus12 wrote: ↑
Mon May 11, 2020 6:40 am
 \((\sqrt{245}-\sqrt{75})^2=\)


A. \(170−5\sqrt8\)

B. \(170−70\sqrt{15}\)

C. \(320−70\sqrt{15}\)

D. \(318−35\sqrt{15}\)

E. \(170\)

[spoiler]OA=C[/spoiler]

Although we can use the formula (x - y)^2 = x^2 - 2xy + y^2, let’s simplify each radical first:

√245 = √49 x √5 = 7√5 and √75 = √25 x √3 = 5√3

So we have:

(7√5 - 5√3)^2 = (7√5)^2 - 2(7√5)(5√3) + (5√3)^2 = 245 - 70√15 + 75 = 320 - 70√15

Answer: C

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