Q1 ) 5+2√3÷7+4√3 = a+b√3
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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
_____ × ______
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.
Answered by
0
Step-by-step explanation:
Given expression
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
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
Subtract 49 from 48 in denominator to get 1.
Now, multiply both term left side to right side.
On, comparing with R.H.S , we have
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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