Question

if root2=1.414,root3=1.732, then find the value of 4/3root3-2root2+3/3root3+2root2

Answer 1

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Answer 2

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

Given,

√2 = 1.414
√3 = 1.732

Now,

$\frac{4}{3 \sqrt{3} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} }$

On rationalizing the denominator we get,

$= \frac{4}{3 \sqrt{3} - 2 \sqrt{2} } \times \frac{3 \sqrt{3} + 2 \sqrt{2} }{3 \sqrt{3} + 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } \times \frac{3 \sqrt{3} - 2 \sqrt{2} }{3 \sqrt{3} - 2 \sqrt{2} } \\ \\ = \frac{4(3 \sqrt{3} + 2 \sqrt{2}) }{ {(3 \sqrt{3} )}^{2} - {(2 \sqrt{2} )}^{2} } + \frac{3(3 \sqrt{3} - 2 \sqrt{2} )}{ {(3 \sqrt{3} })^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} }{27 - 8} + \frac{9 \sqrt{3} - 6 \sqrt{2} }{27 - 8} \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} }{19} + \frac{9 \sqrt{3} - 6 \sqrt{2} }{19} \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} + 9 \sqrt{3} - 6 \sqrt{2} }{19} \\ \\ = \frac{21 \sqrt{3} + 2 \sqrt{2} }{19}$

Putting the values,

$= \frac{21 \times 1.732 + 2 \times 1.414}{19} \\ \\ = \frac{36.372 + 2.828}{19} \\ \\ = \frac{39.2}{19} \\ \\ = 2.06$