Question

SECTION
34) Find the value of a and b , if 3+4 root2 divided by 3-4root 2 =a +b root2

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Answer 1

GIVEN :-

$\\ \sf \: \dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} } = a + b \sqrt{2}$

TO FIND :-

• Value of a and b.

SOLUTION :-

$\\ \sf \: \dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} }$

We will rationalise the denominator.

Rationalising factor is (3 + 4√2).

So, multiplying and dividing numerator and denominator by (3 + 4√2),

$\\ \implies\sf \: \dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} } \times \dfrac{3 + 4 \sqrt{2} }{3 + 4 \sqrt{2} }$

In denominator ,

★ (a-b)(a+b) = a² - b²

• a = 3
• b = 4√2

$\\ \implies \sf \: \dfrac{(3 + {4 \sqrt{2}) }^{2} }{ {3}^{2} - (4 { \sqrt{2} )}^{2} }$

In numerator,

★ (a + b)² = a² + b² + 2ab

• a = 3
• b = 4√2

$\\ \implies \sf \: \dfrac{ {3}^{2} + ( {4 \sqrt{2} ) }^{2} + 2(3)(4 \sqrt{2} ) }{9 - 32} \\ \\ \implies \sf \: \dfrac{9 + 32 + 24 \sqrt{2} }{ - 23} \\ \\ \implies \sf \: \dfrac{41 + 24 \sqrt{2} }{ - 23} \\ \\ \implies \sf \: - \dfrac{ 41}{23} - \dfrac{24 \sqrt{2} }{23}$

This is in the form of a + b√2.

$\\ \implies \sf \: \dfrac{ - 41}{23} - \dfrac{24 \sqrt{2} }{23} = a + b \sqrt{2}$

$\\ \therefore {\underline{\underline{\boxed{ \mathfrak{a = \frac{ - 41}{23} \: \: \: \: \: \: \: \: \: b = \frac{ - 24}{23} }}}}}$

MORE IDENTITIES :-

★ (a-b)² = a² + b² - 2ab

★ (a+x)(a+y) = a² + (x+y)a + xy

★ (a+b)³ = a³ + 3a²b + 3ab² + b³

★ (a-b)³ = a³ - 3a²b + 3ab² - b³