Prove by contradiction that \(\sqrt{3}+\sqrt{5}>\sqrt{11}\). (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
Show Answers Only
\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)
| \(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\) | \(\leqslant\sqrt{11}\) |
| \((\sqrt{3}+\sqrt{5})^2\) | \(\leqslant 11\) |
| \(3+2 \sqrt{15}+5\) | \(\leqslant 11\) |
| \(2 \sqrt{15}\) | \(\leqslant 3\) |
| \(60\) | \(\leqslant 9 \ \text{(incorrect)}\) |
\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)
Show Worked Solution
\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)
| \(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\) | \(\leqslant\sqrt{11}\) |
| \((\sqrt{3}+\sqrt{5})^2\) | \(\leqslant 11\) |
| \(3+2 \sqrt{15}+5\) | \(\leqslant 11\) |
| \(2 \sqrt{15}\) | \(\leqslant 3\) |
| \(60\) | \(\leqslant 9 \ \text{(incorrect)}\) |
\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)