Math, asked by www7, 1 year ago

Q1 ) 5+2√3÷7+4√3 = a+b√3


Answers

Answered by mohammadfaheem
1
5+2√3 7+4√3
_____ × ______
7+4√3 7+4√3

=5(7+4√3)+2√3(7+4√3)
_________________
7×2+2×7×4√3+(4√3)2

=35+20√3 + 14√3+ 8×3
_________________
14+56√3+16×3

=59+34√3
_______
62+56√3

=59+34√3
______
118√3


mysticd: it is wrong.
mysticd: in the first step multiply and divide with conjugate
mysticd: i.e 7-4 root3
www7: yes!!
www7: when we rationalise we change the signs
mohammadfaheem: yeah this one is wrong but please any one delete this answer
mysticd: u can edit the solution
mohammadfaheem: there is no option showing
mohammadfaheem: i think the question is a-b√3
mohammadfaheem: 5+2√3/7+4√3 × 7-4√3/7-4√3 =5(7-4√3)+2√3(7-4√3) / (7)^2-(4√3)^2 =35-20√3+14√3-8×3 / 49-16×3 = 35-20√3+14√3-24 / 49-48 =11-20√3+14√3 / 1 =11-6√3 = a=11 b=6
Answered by Salmonpanna2022
0

Step-by-step explanation:

 \bf{ \underline{Given-}} \\

  \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  =  \bf \: a + b \sqrt{3}  \\

 \bf \underline{To \:  find \:  out-} \\

 \rm{Value \:  of  \:  a \: and \:  b \:  in  \: given \: expression.} \\

 \bf \underline{Solution-} \\

Given expression

  \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  =  \bf \: a + b \sqrt{3}  \\

The denominator is 7 + 4√3.

We know that

Rationalising factor of a + b√c = a - b√c.

So, the rationalising factor of 7 +4√3 = 7-4√3.

On rationalising the denominator them

 \longrightarrow \:   \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  \times  \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\

 \longrightarrow \:  \frac{(5 + 2 \sqrt{3})(7 - 4 \sqrt{3})  }{(7 + 4 \sqrt{3})(7 - 4 \sqrt{3})  }  \\

Now, applying algebraic identity in denominator because it is in the form of;

(a+b)(a-b) = a² - b²

Where, we have to put in our expression: a = 7 and b = 4√3 , we get

\longrightarrow \:  \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2}   } \\

\longrightarrow \:  \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{49 - 48}  \\

Subtract 49 from 48 in denominator to get 1.

\longrightarrow \:  \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{1}  \\

\longrightarrow \: (5 + 2 \sqrt{3} )(7 - 4 \sqrt{3} ) \\

Now, multiply both term left side to right side.

 \longrightarrow \: 35 + 14 \sqrt{3}  - 20 \sqrt{3}  - 8 \sqrt{3 \times 3 } \\

 \longrightarrow \: 35 + 14 \sqrt{3}  - 20 \sqrt{3}  - 24 \\

 \longrightarrow \: (35 - 24) - 6 \sqrt{3}  \\

 \longrightarrow \: 11 - 6 \sqrt{3}  \\

  \bf\therefore \: 11 - 6 \sqrt{3}  = a + b \sqrt{3}  \\

On, comparing with R.H.S , we have

a = 11 and b = -6

  \bf \underline{Hence, value \:  of \: a  = 11 \: and \: b  =  - 6.} \\

Used Formulae:

(a+b)(a-b) = a² - b²

Rationalising factor of a + b√c = a - b√c.

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