Question

The roots of the quadratic equation `"x" + 1/"x" = 3`, x ≠ 0 are:

Options

• `3 + sqrt5, 3 - sqrt5`

• `2 + sqrt5, 2 - sqrt5`

• `(3 + sqrt5)/2, (3 - sqrt5)/2`

• `(3 + sqrt3)/2, (3 - sqrt3)/2`

Answer 1

`(3 + sqrt5)/2, (3 - sqrt5)/2`

Explanation:

We have `"x" + 1/"x"` = 3

⇒ `("x"^2 + 1)/"x"` = 3

⇒ x² + 1 = 3x

On comparing with ax² + bx + c = 0

∴ a = 1, b = –3, c = 1

x = `[(-"b") ± sqrt("b"^2 - 4"ac")]/(2"a")`

= `[-(-3) ± sqrt((-3)^2 - 4(1)(1))]/(2(1))`

= `[3 ± sqrt(9 - 4)]/2`

= `(3 ± sqrt5)/2`
Therefore, `(3 + sqrt5)/2` and `(3 - sqrt5)/2` are the roots of given equation.`