Advertisements
Advertisements
प्रश्न
If `"a"^2 + (1)/"a"^2 = 14`; find the value of `"a" + (1)/"a"`
उत्तर
Using (a + b)2 = a2 + 2ab + b2
`("a" + 1/"a")^2`
= `"a"^2 + 2"a"(1/"a") + (1/"a")^2`
⇒ `("a" + 1/"a")^2 = "a"^2 + 2 + (1)/"a"^2`
⇒ `("a" + 1/"a")^2 = "a"^2 + (1)/"a"^2 + 2`
⇒ `("a" + 1/"a")^2` = 14 + 2
⇒ `("a" + 1/"a")^2` = 16
⇒ `"a" + (1)/"a"` = ±4.
APPEARS IN
संबंधित प्रश्न
Factorise the following:
8a3 – b3 – 12a2b + 6ab2
Write in the expanded form:
(2a - 3b - c)2
Find the value of 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x = 4, y = 3 and z = 2.
Find the following product:
(3x − 4y + 5z) (9x2 +16y2 + 25z2 + 12xy −15zx + 20yz)
Mark the correct alternative in each of the following:
If \[x + \frac{1}{x} = 5\] then \[x^2 + \frac{1}{x^2} = \]
(a − b)3 + (b − c)3 + (c − a)3 =
Find the squares of the following:
(2a + 3b - 4c)
If `"a" + (1)/"a" = 2`, then show that `"a"^2 + (1)/"a"^2 = "a"^3 + (1)/"a"^3 = "a"^4 + (1)/"a"^4`
Find the value of x3 + y3 – 12xy + 64, when x + y = – 4
If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = – 25.