Question

prove that the difference and quotient of 3 + 2 root 3 and 3- 2 root 3 are irrational​

Answer 1

Question:

Prove that the difference and quotient of 3+2√3 and 3-2√3 are irrational.

Solution:

First we will find the difference

$\implies \: 3 + 2 \sqrt{3} - (3 - 2 \sqrt{3} ) \\ \\ = > 3 + 2 \sqrt{3} - 3 + 2 \sqrt{3} \\ \\ = > {4 \sqrt{3} }$

4√3 is irrational because √3 is an irrational number and we know that the product of a non zero rational and an irrational number is always irrational.

Now we will find the quotient

$\implies \frac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } \\ \\ = > \frac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } \times \frac{3 + 2 \sqrt{3} }{3 + 2 \sqrt{3} } \\ \\ = > \frac{ (3 + 2 \sqrt{3} ) {}^{2} }{( 3 ) {}^{2} - (2 \sqrt{3}) {}^{2} } \: \: \\ ( \: by \: using \: identity \: ( {x}^{2} - {y}^{2} ) = (x + y)(x - y) \: ) \\ \\ = > \frac{(3) {}^{2} + (2 \sqrt{3}) {}^{2} + 2(3)(2 \sqrt{3} ) }{9 - 12}$

By using algebraic identity

==> [(a+b)²= a²+b²+2ab]

Now,

$= > \frac{9 + 12 + 12 \sqrt{3} }{ - 3} \\ \\ = > \frac{21 + 12 \sqrt{3} }{ - 3} \\ \\ = > \frac{ - 3( - 7 - 4 \sqrt{3}) }{ - 3} \\ \\ = > - 7 - 4 \sqrt{3}$

-7-4√3 is also an irrational number.

The difference of a rational and irrational number is also an irrational number. Here, 4√3 is irrational.

Hence, the difference and quotient of 3+2√3 and 3-2√3 are irrational.

Hence proved.

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Answer 2

Answer:

Step-by-step explanation: