Question

if a is equals to root 6 + root 5 and b is equals to root 6 minus root 5 then find the value of 2 a square minus 5 ab + 2 b square​

Answer 1

$a = \sqrt{6} + \sqrt{5} \\ \\ \\b = \sqrt{6} - \sqrt{5}$

Now,

$a - b = \sqrt{6} + \sqrt{5}- \sqrt{6} + \sqrt{5} \\ \\ \\ \\ = 2 \sqrt{5}$

Now,

$ab = ( \sqrt{6} + \sqrt{5})( \sqrt{6} - \sqrt{5}) \\ \\ \\ =6 - 5 \\ \\ \\ = 1$

$\rule{200}{2}$

Now ,

$2 {a}^{2} - 5ab + 2 {b}^{2} \\ \\ \\ = 2 {a}^{2}- 4ab + 2 {b}^{2}- ab \\ \\ \\ =2( {a}^{2} - 2ab + {b}^{2})-ab \\ \\ \\ = 2 {(a - b)}^{2}-ab$

Now,

Putting down the values :-

$2 {(2 \sqrt{5})}^{2} - 1 \\ \\ \\ = 2 \times 20- 1 \\ \\ \\ = 40-1 \\ \\ \\ = 39$

$\rule{200}{2}$

Hence ,

The answer is 39

$\rule{200}{2}$

Answer 2

Solution :

If,

a = √6 + √5

b = √6 - √5

Now, for finding the value of (a + b)

(a - b) = √6 + √5 - √6 + √5

=> (a - b) = 2√5

.°. (a - b) = 2√5

Now, for finding the value of (ab)

(ab) = (√6 + √5) ( √6 - √5)

[°.° a - b = (√a + √b)(√a - √b)]

=> (ab) = 6 - 5

=> (ab) = 1

.°. (ab) = 1

Now,

2a^2 - 5ab + 2b^2

= 2a^2 - 4ab - ab + 2b^2

= 2a^2 - 4ab + 2b^2 - ab

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

[°.° (a - b)^2 = a^2 - 2ab + b^2]

= 2(a - b)^2 - ab

Now, putting the value of (a + b).

2(2√5)^2 - 1

= 2 × 20 - 1

= 40 - 1

= 39

.°. 2(a - b)^2 - ab = 39

Answer : 39