Advertisements
Advertisements
प्रश्न
If a + `1/a` = m and a ≠ 0 ; find in terms of 'm'; the value of :
`a - 1/a`
Advertisements
उत्तर
Given that a + `1/a` = m
Now consider the expansion of `( a + 1/a )^2` :
`( a + 1/a )^2 = a^2 + 1/a^2 + 2`
⇒ m2 = a2 + `1/a^2` + 2
⇒ a2 + `1/a^2` = m2 - 2
Now consider the expansion of `( a - 1/a )^2` :
`( a - 1/a )^2 = a^2 + 1/a^2 - 2`
⇒ `( a - 1/a )^2 = m^2 - 2 - 2`
⇒ `( a - 1/a )^2 = m^2 - 4`
⇒ `( a - 1/a ) = +-sqrt(m^2 - 4)`
APPEARS IN
संबंधित प्रश्न
Expand: `( 2x - 1/x )( 3x + 2/x )`
Expand : ( 5x - 3y - 2 )2
If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.
If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.
If x+ y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.
If x > 0 and `x^2 + 1/[9x^2] = 25/36, "Find" x^3 + 1/[27x^3]`
If x2 + `x^(1/2)`= 7 and x ≠ 0; find the value of:
7x3 + 8x − `7/x^3 - 8/x`
The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : The sum of their squares
Find the value of 'a': 4x2 + ax + 9 = (2x + 3)2
If x = `1/( x - 5 ) "and x ≠ 5. Find" : x^2 - 1/x^2`
