Advertisements
Advertisements
प्रश्न
If x = `(4 - sqrt(15))`, find the values of:
`(x + (1)/x)^2`
Advertisements
उत्तर
x = `4 - sqrt(15)`
∴ `(1)/x = (1)/(4 - sqrt(15))`
= `(1)/(4 - sqrt(15)) xx (4 + sqrt(15))/(4 + sqrt(15))`
= `(4 + sqrt(15))/(4^2 - (sqrt(15))^2`
= `(4 + sqrt(15))/(16 - 15)`
= `(4 + sqrt(15))/(1)`
= `4 + sqrt(15)`
∴ `x + (1)/x = (4 - sqrt(15)) + (4 + sqrt(15))`
= `4 - sqrt(15) + 4 + sqrt(15)`
= 8
Hence, `(x + (1)/x)^2`
= (8)2
= 64
APPEARS IN
संबंधित प्रश्न
Rationalize the denominator.
`12/(4sqrt3 - sqrt 2)`
Rationalise the denominators of : `3/sqrt5`
Rationalise the denominators of:
`[sqrt6 - sqrt5]/[sqrt6 + sqrt5]`
Simplify :
` 22/[2sqrt3 + 1] + 17/[ 2sqrt3 - 1]`
Rationalise the denominator of `1/[ √3 - √2 + 1]`
Simplify by rationalising the denominator in the following.
`(42)/(2sqrt(3) + 3sqrt(2)`
If `(sqrt(2.5) - sqrt(0.75))/(sqrt(2.5) + sqrt(0.75)) = "p" + "q"sqrt(30)`, find the values of p and q.
If x = `(7 + 4sqrt(3))`, find the value of
`sqrt(x) + (1)/(sqrt(x)`
If x = `(1)/((3 - 2sqrt(2))` and y = `(1)/((3 + 2sqrt(2))`, find the values of
x2 + y2
Using the following figure, show that BD = `sqrtx`.
