Advertisements
Advertisements
प्रश्न
In the following, find the value of a and b:
`(7 + sqrt(5))/(7 - sqrt(5)) - (7 - sqrt(5))/(7 + sqrt(5)) = "a" + "b"sqrt(5)`
उत्तर
`(7 + sqrt(5))/(7 - sqrt(5)) - (7 - sqrt(5))/(7 + sqrt(5)`
= `(7 + sqrt(5))/(7 - sqrt(5)) xx (7 + sqrt(5))/(7 + sqrt(5)) - (7 - sqrt(5))/(7 + sqrt(5)) xx (7 - sqrt(5))/(7 - sqrt(5))`
= `((7 + sqrt(5))^2)/(7^2 - (sqrt(5))^2) - (7 - sqrt(5))^2/(7 ^2 - (sqrt(5))^2`
= `(7^2 + 2 xx 7 xx sqrt(5) + (sqrt(5))^2)/(49 - 5) - (7^2 - 2 xx 7 xx sqrt(5) + (sqrt(5))^2)/(49 - 5)`
= `(49 + 14sqrt(5) + 5)/(44) - (49 - 14sqrt(5) + 5)/(44)`
= `(54 + 14sqrt(5))/(44) - (54 - 14sqrt(5))/(44)`
= `(2(27 + 7sqrt(5)))/(44) - (2(22 - 7sqrt(5)))/(44)`
= `(27 + 7sqrt(5))/(22) - (27 - 7sqrt(5))/(22)`
= `(27)/(22) + (7sqrt(5))/(22) - (27)/(22) + (7sqrt(5))/(22)`
= `(14sqrt(5))/(22)`
= `(7sqrt(5))/(11)`
= `0 + (7sqrt(5))/(11)`
= `"a" + "b"sqrt(5)`
Hence, a = 0 and b = `(7)/(11)`.
APPEARS IN
संबंधित प्रश्न
Rationalise the denominators of : `1/(sqrt3 - sqrt2 )`
Rationalise the denominators of : `[ 2 - √3 ]/[ 2 + √3 ]`
Simplify by rationalising the denominator in the following.
`(42)/(2sqrt(3) + 3sqrt(2)`
Simplify by rationalising the denominator in the following.
`(sqrt(3) + 1)/(sqrt(3) - 1)`
Simplify by rationalising the denominator in the following.
`(3 - sqrt(3))/(2 + sqrt(2)`
In the following, find the values of a and b:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = "a" + "b"sqrt(3)`
In the following, find the values of a and b:
`(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)) = "a" - "b"sqrt(6)`
Evaluate, correct to one place of decimal, the expression `5/(sqrt20 - sqrt10)`, if `sqrt5` = 2.2 and `sqrt10` = 3.2.
Draw a line segment of length `sqrt8` cm.
Show that: `x^3 + 1/x^3 = 52`, if x = 2 + `sqrt3`
