Question

If
$\rm \dfrac{\sqrt{28-10\sqrt{3}}+\sqrt{7+4\sqrt{3}}}{ \sqrt{16+6\sqrt{7}}} = a + b \sqrt{7}$
then what is the value of (2a+b)?​

Answer 1

Solution :-

Given that

[√(28-10√3)+√(7+4√3)]/[√(16+6√7)] = a+b√7 ----(1)

√(28-10√3)

= √(28-2×5√3)

= √[28-2√(3×25)]

= √(28-2√75)

=√[(25+3)-2√(25×3)]

= √[(√25)²+(√3)²-2(√25)(√3)]

It is in the form of a²+b²-2ab

We know that

(a-b)² = a²-2ab+b²

Therefore, √(28-10√3) = [√(√25-√3)]²

=> √(28-10√3) = √25 -√3

=> √(28-10√3) = 5-√3 ----------(2)

and

√(7+4√3)

= √[(4+3)+2√(4×3)]

=> √[(√4)²+(√3)²+2(√4)(√3)]

It is in the form of a²+b²+2ab

We know that

(a+b)² = a²+2ab+b²

Therefore, √(7+4√3) = [√(√4+√3)]²

=> √(7+4√3) = √4+√3

=> √(7+4√3) = 2+√3 ------------(3)

and

√(16+6√7)

=√[(9+7)+2(√7×9)]

= √[(√9)²+(√7)²+2(√9)(√7)]

It is in the form of a²+b²+2ab

We know that

(a+b)² = a²+2ab+b²

Therefore, √(16+6√7) = [(√9)+√7)]²

=> √(16+6√7) = √9+√7

=> √(16+6√7) = 3+√7 ----------(4)

Now, (1) becomes

[(5-√3)+(2+√3)]/(3+√7) = a+b√7

=> (5-√3+2+√3)/(3+√7) = a+b√7

=> (5+2)/(3+√7) = a+b√7

=> 7/(3+√7) = a+b√7

We know that

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

The Rationalising factor of 3+√7 is 3-√7

On Rationalising the denominator then

=> [7/(3+√7)]×[(3-√7)/(3-√7)] = a+b√7

=> [7(3-√7)]/[(3+√7)(3-√7)] = a+b√7

=> [7(3-√7)]/[3²-(√7)²] = a+b√7

Since, (a+b)(a-b) = a²-b²

=> [7(3-√7)]/(9-7) = a+b√7

=> 7(3-√7)/2 = a+b√7

=> (21-7√7)/2 = a+b√7

=> (21/2)-(7/2)√7 = a+b√7

=> (21/2)+(-7/2)√7 = a+b√7

On comparing both sides then

a = 21/2

b = -7/2

Now, The value of 2a+b

=> 2(21/2)+(-7/2)

=> 21-(7/2)

=> (42-7)/2

=> 35/2

Answer :-

The value of 2a+b is 35/2

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ (a-b)² = a²-2ab+b²

→ (a+b)(a-b) = a²-b²

→ The Rationalising factor of a+√b = a-√b

Answer 2

Step-by-step explanation:

Solution :-

Given that

[√(28-10√3)+√(7+4√3)]/[√(16+6√7)] = a+b√7 ----(1)

√(28-10√3)

= √(28-2×5√3)

= √[28-2√(3×25)]

= √(28-2√75)

=√[(25+3)-2√(25×3)]

= √[(√25)²+(√3)²-2(√25)(√3)]

It is in the form of a²+b²-2ab

We know that

(a-b)² = a²-2ab+b²

Therefore, √(28-10√3) = [√(√25-√3)]²

=> √(28-10√3) = √25 -√3

=> √(28-10√3) = 5-√3 ----------(2)

and

√(7+4√3)

= √[(4+3)+2√(4×3)]

=> √[(√4)²+(√3)²+2(√4)(√3)]

It is in the form of a²+b²+2ab

We know that

(a+b)² = a²+2ab+b²

Therefore, √(7+4√3) = [√(√4+√3)]²

=> √(7+4√3) = √4+√3

=> √(7+4√3) = 2+√3 ------------(3)

and

√(16+6√7)

=√[(9+7)+2(√7×9)]

= √[(√9)²+(√7)²+2(√9)(√7)]

It is in the form of a²+b²+2ab

We know that

(a+b)² = a²+2ab+b²

Therefore, √(16+6√7) = [(√9)+√7)]²

=> √(16+6√7) = √9+√7

=> √(16+6√7) = 3+√7 ----------(4)

Now, (1) becomes

[(5-√3)+(2+√3)]/(3+√7) = a+b√7

=> (5-√3+2+√3)/(3+√7) = a+b√7

=> (5+2)/(3+√7) = a+b√7

=> 7/(3+√7) = a+b√7

We know that

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

The Rationalising factor of 3+√7 is 3-√7

On Rationalising the denominator then

=> [7/(3+√7)]×[(3-√7)/(3-√7)] = a+b√7

=> [7(3-√7)]/[(3+√7)(3-√7)] = a+b√7

=> [7(3-√7)]/[3²-(√7)²] = a+b√7

Since, (a+b)(a-b) = a²-b²

=> [7(3-√7)]/(9-7) = a+b√7

=> 7(3-√7)/2 = a+b√7

=> (21-7√7)/2 = a+b√7

=> (21/2)-(7/2)√7 = a+b√7

=> (21/2)+(-7/2)√7 = a+b√7

On comparing both sides then

a = 21/2

b = -7/2

Now, The value of 2a+b

=> 2(21/2)+(-7/2)

=> 21-(7/2)

=> (42-7)/2

=> 35/2

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