What will be the rationalizing term for $30\sqrt 2 + 24 + 20 \sqrt 6$? It cannot be $30 \sqrt 2 - 20\sqrt 6$. Neither can it be $30 \sqrt 2+ 20 \sqrt 6 - 24$ because the product of these with the denominator will again contain a square root. Please help. Thank you.
2026-03-26 10:07:09.1774519629
Rationalising term
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1
Take $(30\sqrt2+24+20\sqrt6)\times(a+b\sqrt2+c\sqrt3+d\sqrt6)$ $=30a\sqrt2+60b+30c\sqrt6+60d\sqrt3+24a+24b\sqrt2+24c\sqrt3+24d\sqrt6+20a\sqrt6+40b\sqrt3+60c\sqrt2+120d$ $=(60b+24a+120d)+(30a+24b+60c)\sqrt2+(60d+24c+40b)\sqrt3+(30c+24d+20a)\sqrt6$
From this, we get $$5a+4b+10c=0$$ $$15d+6c+10b=0$$ $$15c+12d+10a=0$$ Since we have $4$ variables and $3$ equations, we need to assume some value. Say $a=1$ $$b={{245}\over{604}}, c={{-100}\over{151}}, b={{-5}\over{906}}$$ Since multiplying won't affect rationalisation, let us multiply all 4 variables by $1812$. $$a=1812, b=735, c=-1200, d=-10$$ Therefore, your rationalising factor is $(1812+735\sqrt2-1200\sqrt3-10\sqrt6)$