Question

please answer this
it's urgent if you don't know please don't answer​

Answer 1

Step-by-step explanation:

Given :-

a = (√3+1)/(√3-1)

b = (√3-1)/(√3+1)

To find :-

Find the value of a^2+ab+b^2 ?

Solution:-

Given that

a = (√3+1)/(√3-1)

The denominator = √3-1

The Rationalising factor of√3-1 is √3+1

On Rationalising the denominator then

=> a = [(√3+1)/(√3-1)]×[(√3+1)/(√3+1)]

=> a =[(√3+1)(√3+1)]/[(√3-1)(√3+1)]

=> a = (√3+1)^2/[(√3)^2-(1)^2]

Since (a+b)(a-b)=a^2-b^2

=>a = (√3+1)^2/(3-1)

=> a = (√3+1)^2/2

=> a = [(√3)^2+2(√3)(1)+(1)^2]/2

Since (a+b)^2 = a^2+2ab+b^2

=> a = (3+1+2√3)/2

=> a = (4+2√3)/2

=> a = 2(2+√3)/2

=> a = 2+√3 ---------------(1)

and

On squaring both sides then

=> a^2 = (2+√3)^2

=> a^2=2^2+(√3)^2+2(2)(√3)

Since (a+b)^2 = a^2+2ab+b^2

=> a^2 = 4+3+2√6

=> a^2 = 7+2√6 -----------(2)

and

b = (√3-1)/(√3+1)

The denominator = √3+1

The Rationalising factor of√3+1 is √3-1

On Rationalising the denominator then

=> b = [(√3-1)/(√3+1)]×[(√3-1)/(√3-1)]

=> b =[(√3-1)(√3-1)]/[(√3+1)(√3-1)]

=> b = (√3-1)^2/[(√3)^2-(1)^2]

Since (a+b)(a-b)=a^2-b^2

=>b = (√3-1)^2/(3-1)

=> b = (√3-1))^2/2

=> b = [(√3)^2-2(√3)(1)+(1)^2]/2

Since (a-b)^2 = a^2-2ab+b^2

=> b = (3+1-2√3)/2

=> b = (4-2√3)/2

=> b = 2(2-√3)/2

=> b= 2-√3 ---------------(3)

and

On squaring both sides then

=> b^2 = (2-√3)^2

=> b^2=2^2+(√3)^2-2(2)(√3)

Since (a-b)^2 = a^2-2ab+b^2

=> b^2 = 4+3-2√6

=> b^2 = 7-2√6 -----------(4)

Now , a^2+ab+b^2

From (1),(2),(3)&(4)

=> (7+2√6)+(2+√3)(2-√3)+(7-2√6)

=> 7+2√6+(2^2-(√3)^2)+7-2√6

=> 7+2√6+(4-3)+7-2√6

=> 7+2√6+1+7-2√6

=> 7+1+7

=> 15

Answer:-

The value of a^+ab+b^2 for the given problem is 15

Used formulae:-

• (a+b)(a-b)=a^2-b^2

• (a-b)^2 = a^2-2ab+b^2

• (a+b)^2 = a^2+2ab+b^2

• The Rationalising factor of a+√b is a-√b

• The Rationalising factor of a-√b is a+√b

Answer 2

Answer:

15 is the required answer

Step-by-step explanation:

${\small{\mathtt{a = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } }}} \\{\small{\mathtt{a = \frac{ {( \sqrt{3} + 1) }^{2} }{ {( \sqrt{3} )}^{2} - {1}^{2} } = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} }}} \\{\small{\mathtt{ a = \frac{4 + 2 \sqrt{3} }{2} = 2 + \sqrt{3} }}}$

${\small{\mathtt{b = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1 }}}} \\ {\small{\mathtt{b = \frac{ {( \sqrt{3 } - 1 )}^{2} }{ {( \sqrt{3} )}^{2} - 1 } = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1}}}} \\ {\small{\mathtt{b = \frac{4 - 2 \sqrt{3} }{2} = 2 - \sqrt{3} }}}$

${\small{\mathtt{a² + b² + ab=(2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}+(2+\sqrt{3})(2-\sqrt{3})}}} \\ {\small{\mathtt{= {2}^{2} + { (\sqrt{3} )}^{2} + 4 \sqrt{3} + {2}^{2} + { (\sqrt{3} )}^{2} - 4 \sqrt{3} + {2}^{2} - { (\sqrt{3} )}^{2} }}} \\ {\small{\mathtt{ = 4 + 3 {\cancel{+ 4 \sqrt{3} } + 4 + 3 - {\cancel{ 4 \sqrt{3}}}} + 4 - 3 }}}\\ {\small{\mathtt{= 7 + 7 + 1}}} \\ {\small{\mathtt{\red{= 15}}}}$

${\small{\boxed{\mathcal{\purple{a²+b²+ab=15}}}}}ans$