Question

The roots of the quadratic equation `2"x"^2 - 2sqrt2"x" + 1 = 0` are:

Options

• `1/sqrt2, 1/sqrt2`

• `sqrt2, sqrt2`

• `1/sqrt2, -1/sqrt2`

• `sqrt2, 1/sqrt2`

Answer 1

`1/sqrt2, 1/sqrt2`

Explanation:

We have `2"x"^2 - 2sqrt2"x" + 1 = 0`

Here a = 2, b = `-2sqrt2`, c = 1

x = `(-"b" ± sqrt"D")/(2"a")`

= `(-(-2sqrt2) ± 0)/(2 xx 2)`

= `(2sqrt2)/4`

= `sqrt2/2`

= `1/sqrt2`

∴ x = `1/sqrt2, 1/sqrt2`