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If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b},\) where \(a\) and \(b\) are integers and

Expert replies
by VJesus12 » Sun Jun 14, 2020 7:31 am

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E

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If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b},\) where \(a\) and \(b\) are integers and \(b < a,\) then \(a - b =\)

A. 1
B. 2
C. 3
D. 4
E. 5

[spoiler]OA=E[/spoiler]

Source: Magoosh
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Source: — Problem Solving |

VJesus12 wrote:
Sun Jun 14, 2020 7:31 am
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b},\) where \(a\) and \(b\) are integers and \(b < a,\) then \(a - b =\)

A. 1
B. 2
C. 3
D. 4
E. 5

[spoiler]OA=E[/spoiler]

Source: Magoosh
So, we have \(\sqrt{17+\sqrt{264}}=\sqrt{a}+\sqrt{b}\)

Squaring both sides, we get

\(17 + \sqrt{264} = (\sqrt{a}+\sqrt{b})^2\)

\(17 + \sqrt{264} = (a+b+2\sqrt{ab})\)

=> \(a+b = 17\) and \(2\sqrt{ab}=\sqrt{264}=> 4ab=264 => ab = 66\)

So, we have

\(a+b=17\) and
\(ab=66\)

By hit and trial, we can get that \(a=11\) and \(b=6\); thus, \(a-b=11-6=5\).

Correct answer: E

Hope this helps!

-Jay
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VJesus12 wrote:
Sun Jun 14, 2020 7:31 am
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b},\) where \(a\) and \(b\) are integers and \(b < a,\) then \(a - b =\)

A. 1
B. 2
C. 3
D. 4
E. 5

[spoiler]OA=E[/spoiler]

Solution:

We are given that:

√(17 + √264) = √a + √b

Squaring both sides, we have:

17 + √264 = a + b + 2√(ab)

17 + 2√66 = a + b + 2√(ab)

We see that a + b must be 17 and ab must be 66. That is:

a + b = 17 and ab = 66

We can solve these two equations algebraically, but since we are given that a and b are integers, we see that if a = 11 and b = 6, the two equations will be satisfied. In that case, we have a - b = 11 - 6 = 5.

Answer: E

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