Find the value of m and n: if:
3+23−2=m+n2\dfrac{3 + \sqrt2}{3 - \sqrt2} = m + n \sqrt23−23+2=m+n2
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3+23−2=3+23−2×3+23+2=(3+2)×(3+2)(3−2)×(3+2)=(3+2)2(3)2−(2)2=9+2+629−2=11+627=117+627\dfrac{3 + \sqrt2}{3 - \sqrt2} = \dfrac{3 + \sqrt2}{3 - \sqrt2} \times \dfrac{3 + \sqrt2}{3 + \sqrt2}\\[1em] = \dfrac{(3 + \sqrt2) \times (3 + \sqrt2)}{(3 - \sqrt2) \times (3 + \sqrt2)}\\[1em] = \dfrac{(3 + \sqrt2)^2}{(3)^2 - (\sqrt2)^2}\\[1em] = \dfrac{9 + 2 + 6\sqrt2}{9 - 2}\\[1em] = \dfrac{11 + 6\sqrt2}{7}\\[1em] = \dfrac{11}{7} + \dfrac{6\sqrt2}{7}3−23+2=3−23+2×3+23+2=(3−2)×(3+2)(3+2)×(3+2)=(3)2−(2)2(3+2)2=9−29+2+62=711+62=711+762
Given :3+23−2=m+n2\dfrac{3 + \sqrt2}{3 - \sqrt2} = m + n \sqrt23−23+2=m+n2
⇒ 117+627=m+n2\dfrac{11}{7} + \dfrac{6\sqrt2}{7} = m + n \sqrt2711+762=m+n2
Hence, m = 117\dfrac{11}{7}711 and n = 67\dfrac{6}{7}76.
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Simplify :
5−105+10−5+105−10\dfrac{5 - \sqrt{10}}{5 + \sqrt{10}} - \dfrac{5 + \sqrt{10}}{5 - \sqrt{10}}5+105−10−5−105+10
717−23−317+23\dfrac{7}{\sqrt{17} - 2\sqrt{3}} - \dfrac{3}{\sqrt{17} + 2\sqrt{3}}17−237−17+233
5+237+43=m+n3\dfrac{5 + 2\sqrt3}{7 + 4\sqrt3} = m + n\sqrt37+435+23=m+n3
By rationalising the denominator of each of the following; find, in each case, the value correct to two significant figures :
(i) 13−2\dfrac{1}{3 - \sqrt2}3−21
(ii) 12+3\dfrac{1}{2 + \sqrt3}2+31
(iii) 432−23\dfrac{4}{3\sqrt2 - 2\sqrt3}32−234
