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If p = \sqrt3 -2/\sqrt2 + 1, then which of the following
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16 Apr 2019, 01:26
16 Apr 2019, 01:26
Expert Reply
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Question Stats:
55% (01:14) correct
44% (01:51) wrong based on 9 sessions
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If \(p = \frac{\sqrt{3} -2}{\sqrt{2} + 1}\), then which of the following equals \(p-4\) ?
A. \(\sqrt{3} - 2\)
B. \(\sqrt{3} + 2\)
C. \(2\)
D. \(-2 \sqrt{2} + \sqrt{6} - \sqrt{3} - 2\)
E. \(- 2 \sqrt{2} + \sqrt{6} - \sqrt{3} + 2\)
A. \(\sqrt{3} - 2\)
B. \(\sqrt{3} + 2\)
C. \(2\)
D. \(-2 \sqrt{2} + \sqrt{6} - \sqrt{3} - 2\)
E. \(- 2 \sqrt{2} + \sqrt{6} - \sqrt{3} + 2\)
Re: If p = \sqrt3 -2/\sqrt2 + 1, then which of the following
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09 Nov 2019, 15:59
09 Nov 2019, 15:59
any explanation on this one algebraically? can do it via calculator easily but not the point...
Re: If p = \sqrt3 -2/\sqrt2 + 1, then which of the following
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09 Nov 2019, 19:05
09 Nov 2019, 19:05
1
Expert Reply
The question is easy if you do know the exponent rule of rationalization
\(p= \frac{\sqrt{3} - 2 }{\sqrt{2} + 1 }\) \(= \frac{\sqrt{3} - 2}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} -1} =\) \(\frac{\sqrt{3} \sqrt{2} + \sqrt{3} -1 + -2 \sqrt{2} + -2-1}{(\sqrt{2})^2 -1^2}\)
Do the calculation
\(p-4=\sqrt{6} - \sqrt{3}-2 \sqrt{2} + 2= \sqrt{6} - \sqrt{3} -2 \sqrt{2} -2 = D\)
\(p= \frac{\sqrt{3} - 2 }{\sqrt{2} + 1 }\) \(= \frac{\sqrt{3} - 2}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} -1} =\) \(\frac{\sqrt{3} \sqrt{2} + \sqrt{3} -1 + -2 \sqrt{2} + -2-1}{(\sqrt{2})^2 -1^2}\)
Do the calculation
\(p-4=\sqrt{6} - \sqrt{3}-2 \sqrt{2} + 2= \sqrt{6} - \sqrt{3} -2 \sqrt{2} -2 = D\)
