Question

In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

`2x^2 - 2sqrt2x + 1 = 0`

Answer 1

We have been given, `2x^2-2sqrt2x+1=0`

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:

D = b² - 4ac

Now, according to the equation given to us, we have, a = 2, `b=-2sqrt2` and c = 1.

Therefore, the discriminant is given as,

`D=(-2sqrt2)^2-4(2)(1)`

= 8 - 8

= 0

Since, in order for a quadratic equation to have real roots, D ≥ 0. Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(-2sqrt2)+-sqrt0)/(2(2))`

`=(2sqrt2)/4`

`=1/sqrt2`

Therefore, the roots of the equation are real and equal and its value is `1/sqrt2`.