Question

if the real positive fourth root of(28+16√3) is divided by (2+√3) and the result is expressed as (a+b√3) with a&b both integers then (4a+3b) is equal to​

Answer 1

SOLUTION

GIVEN

The real positive fourth root of (28+16√3) is divided by (2+√3) The result is expressed as (a+b√3) with a & b both integers

TO DETERMINE

The value of (4a+3b)

EVALUATION

Here

$\sf{}28 + 16 \sqrt{3}$

$= \sf{}16 + 12 + (2 \times 4 \times 2 \sqrt{3} )$

$= \sf{} {(4)}^{2} + {(2 \sqrt{3}) }^{2} + (2 \times 4 \times 2 \sqrt{3} )$

$= \sf{} {(4 + 2 \sqrt{3} )}^{2}$

So

$\sf{} 28 + 16 \sqrt{3} = {(4 + 2 \sqrt{3} )}^{2}$

Since we are treating with real positive fourth root of (28+16√3)

$\therefore \: \sf{} \sqrt{ 28 + 16 \sqrt{3} } = {4 + 2 \sqrt{3} }$

Now

$\sf{}4 + 2 \sqrt{3}$

$= \sf{}3 + 1 + 2 \sqrt{3}$

$= \sf{} {( \sqrt{3}) }^{2} + {(1)}^{2} + 2 \times 1 \times \sqrt{3}$

$= \sf{} {( \sqrt{3} + 1)}^{2}$

$\therefore \: \sf{} \sqrt[4]{28 + 16 \sqrt{3}} = \sqrt{3} + 1$

So

$\displaystyle \sf{} \frac{ \sqrt[4]{28 + 16 \sqrt{3} } }{2 + \sqrt{3} }$

$= \displaystyle \sf{} \frac{ \sqrt{3} + 1 }{2 + \sqrt{3} }$

$= \displaystyle \sf{} \frac{2( \sqrt{3} + 1) }{4 +2 \sqrt{3} }$

$= \displaystyle \sf{} \frac{2( \sqrt{3} + 1) }{ {( \sqrt{3} + 1)}^{2} }$

$= \displaystyle \sf{} \frac{2 }{ {( \sqrt{3} + 1)} }$

$= \displaystyle \sf{} \frac{2 ( \sqrt{3} - 1) }{ {( \sqrt{3} + 1)( \sqrt{3} - 1)} }$

$= \displaystyle \sf{} \frac{2 ( \sqrt{3} - 1) }{ {( 3 - 1)} }$

$= \displaystyle \sf{} \frac{2 ( \sqrt{3} - 1) }{ {2} }$

$\sf{} = \sqrt{3} - 1$

$\sf{} = - 1 + \sqrt{3}$

$\sf{}Which \: is \: of \: the \: form \: \: a + b \sqrt{3}$

Where a = - 1 and b = 1

Hence

$\sf{}4a + 3b$

$= \sf{} - 4 + 3$

$= \sf{} - 1$

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If x=√3a+2b + √3a-2b / √3a+2b - √3a-2b

prove that bx²-3ax+b=0

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Answer 2

Step-by-step explanation:

Given if the real positive fourth root of(28+16√3) is divided by (2+√3) and the result is expressed as (a+b√3) with a&b both integers then (4a+3b) is equal to

• We have √x + √y = √x + √x^2 – y / 2 + √x - √x^2 – y / 2
• We have x = 28, y = 16^2 x 3
•                             = 768
• Applying the identity we get
•          So √28 + √768 = √28 + √16 / 2 + √28 - √16 / 2
•                                    = √16 + √12
•                                    = 4 + 2√3
• Now a = 4, b = 4 x 3 = 12
• So we have
•                        √4 + √12 = √4 + √4^2 – 12 / 2  +  √4 - √4^2 – 12 / 2
•                                       = √4 + 2 / 2 + √4 – 2 / 2
•                                       = √3 + 1
• So the fourth root is √3 + 1 and this is divided by 2 + √3 we get
•                    So 1 + √3 / 2 + √3
• So rationalizing the denominator we get
•                      So 1 + √3 / 2 + √3 x 2 - √3 / 2 - √3
•                           (1 + √3 ) (2 - √3) / 4 – 3
•                              2 + √3 – 3
•                          So – 1 + 1 √3
• We have a = - 1, b = 1
•                       So 4a + 3b = 4 (- 1) + 3(1)
•                                       = - 4 + 3
•                                       = - 1

Reference link will be

https://brainly.in/question/15882667