Advertisements
Advertisements
Question
Rationalize the denominator.
`4/(7+ 4 sqrt3)`
Advertisements
Solution
`4/(7+ 4 sqrt3)`
`= 4/(7+ 4 sqrt3) xx (7- 4 sqrt3)/(7 - 4 sqrt3) `
`= (4(7- 4 sqrt3))/((7)^2 - (4 sqrt 3)^2)` ....`[(a + b)(a - b) = a^2 - b^2]`
`= (4(7- 4 sqrt3)) /(49 - 48)`
`= 28 - 16 sqrt 3`
APPEARS IN
RELATED QUESTIONS
Rationalize the denominator.
`1/(sqrt 7 + sqrt 2)`
Rationalize the denominator.
`3/(2 sqrt 5 - 3 sqrt 2)`
Simplify by rationalising the denominator in the following.
`(3sqrt(2))/sqrt(5)`
Simplify by rationalising the denominator in the following.
`(1)/(sqrt(3) + sqrt(2))`
Simplify by rationalising the denominator in the following.
`(42)/(2sqrt(3) + 3sqrt(2)`
Simplify by rationalising the denominator in the following.
`(7sqrt(3) - 5sqrt(2))/(sqrt(48) + sqrt(18)`
Simplify the following
`(sqrt(7) - sqrt(3))/(sqrt(7) + sqrt(3)) - (sqrt(7) + sqrt(3))/(sqrt(7) - sqrt(3)`
Simplify the following :
`sqrt(6)/(sqrt(2) + sqrt(3)) + (3sqrt(2))/(sqrt(6) + sqrt(3)) - (4sqrt(3))/(sqrt(6) + sqrt(2)`
Simplify the following :
`(3sqrt(2))/(sqrt(6) - sqrt(3)) - (4sqrt(3))/(sqrt(6) - sqrt(2)) + (2sqrt(3))/(sqrt(6) + 2)`
In the following, find the values of a and b.
`(sqrt(3) - 1)/(sqrt(3) + 1) = "a" + "b"sqrt(3)`
In the following, find the values of a and b:
`(1)/(sqrt(5) - sqrt(3)) = "a"sqrt(5) - "b"sqrt(3)`
If x = `(4 - sqrt(15))`, find the values of
`x + (1)/x`
If x = `((2 + sqrt(5)))/((2 - sqrt(5))` and y = `((2 - sqrt(5)))/((2 + sqrt(5))`, show that (x2 - y2) = `144sqrt(5)`.
If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) + 1)`, find the values of
x3 + y3
If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) - 1)`, find the values of
x2 - y2 + xy
If x = `(1)/((3 - 2sqrt(2))` and y = `(1)/((3 + 2sqrt(2))`, find the values of
x2 + y2
If x = `sqrt3 - sqrt2`, find the value of:
(i) `x + 1/x`
(ii) `x^2 + 1/x^2`
(iii) `x^3 + 1/x^3`
(iv) `x^3 + 1/x^3 - 3(x^2 + 1/x^2) + x + 1/x`
Draw a line segment of length `sqrt5` cm.
Show that: `x^2 + 1/x^2 = 34,` if x = 3 + `2sqrt2`
