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\frac{6+4 \sqrt{3}}{2+\sqrt{3}} =
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25 Jul 2026, 07:41
25 Jul 2026, 07:41
1
Expert Reply
00:00
Question Stats:
71% (01:24) correct
28% (03:23) wrong based on 7 sessions
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\( \frac{6+4 \sqrt{3}}{2+\sqrt{3}} = \)
A. \(2-\sqrt{3}\)
B. \(2 \sqrt{3}\)
C. \(2+\sqrt{3}\)
D. \(4-2 \sqrt{3}\)
E. \(4+2 \sqrt{3}\)
A. \(2-\sqrt{3}\)
B. \(2 \sqrt{3}\)
C. \(2+\sqrt{3}\)
D. \(4-2 \sqrt{3}\)
E. \(4+2 \sqrt{3}\)
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
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\frac{6+4 \sqrt{3}}{2+\sqrt{3}} =
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25 Jul 2026, 23:45
25 Jul 2026, 23:45
Expert Reply
Explanation
Use calculator here
\(\frac{6+4(1.73)}{2+1.73}=?\)
\(\frac{6+6.92}{3.73}\) nearly equals \(3.5\)
Find a value which is closest to 3.5
Only B equals
Answer: B
Use calculator here
\(\frac{6+4(1.73)}{2+1.73}=?\)
\(\frac{6+6.92}{3.73}\) nearly equals \(3.5\)
Find a value which is closest to 3.5
Only B equals
Answer: B
Re: \frac{6+4 \sqrt{3}}{2+\sqrt{3}} =
[#permalink]
26 Jul 2026, 04:00
26 Jul 2026, 04:00
Expert Reply
\({Step-by-Step Solution}\)
Step 1 - Recognize that the denominator contains a radical
We need to rationalize the denominator by multiplying numerator and denominator by the conjugate \(2-\sqrt{3}\) .
\(\frac{6+4 \sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)
Step 2 - Multiply the numerators
\(\begin{gathered}
(6+4 \sqrt{3})(2-\sqrt{3})=6(2)-6(\sqrt{3})+4 \sqrt{3}(2)-4 \sqrt{3}(\sqrt{3}) \\
=12-6 \sqrt{3}+8 \sqrt{3}-4(3) \\
=12-12+2 \sqrt{3}=2 \sqrt{3}
\end{gathered}\)
Step 3 - Multiply the denominators
\((2+\sqrt{3})(2-\sqrt{3})=4-3=1\)
Step 4 - Simplify the fraction
\(\begin{gathered}
\frac{2 \sqrt{3}}{1}=2 \sqrt{3} \\
2 \sqrt{3}
\end{gathered}\)
Step 1 - Recognize that the denominator contains a radical
We need to rationalize the denominator by multiplying numerator and denominator by the conjugate \(2-\sqrt{3}\) .
\(\frac{6+4 \sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)
Step 2 - Multiply the numerators
\(\begin{gathered}
(6+4 \sqrt{3})(2-\sqrt{3})=6(2)-6(\sqrt{3})+4 \sqrt{3}(2)-4 \sqrt{3}(\sqrt{3}) \\
=12-6 \sqrt{3}+8 \sqrt{3}-4(3) \\
=12-12+2 \sqrt{3}=2 \sqrt{3}
\end{gathered}\)
Step 3 - Multiply the denominators
\((2+\sqrt{3})(2-\sqrt{3})=4-3=1\)
Step 4 - Simplify the fraction
\(\begin{gathered}
\frac{2 \sqrt{3}}{1}=2 \sqrt{3} \\
2 \sqrt{3}
\end{gathered}\)
