In the following, determine whether the given quadratic equation

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`

Page 2

We have been given, 3x2 - 5x + 2 = 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 = 3, b = -5 and c = 2.

Therefore, the discriminant is given as,

D = (-5)2 - 4(3)(2)

= 25 - 24

= 1

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 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=(-(-5)+-sqrt1)/(2(3))`

`=(5+-1)/6`

Now we solve both cases for the two values of x. So, we have,

`x=(5+1)/6`

`=6/6`

= 1

Also,

`x=(5-1)/6`

`=4/6`

`=2/3`

Therefore, the roots of the equation are `2/3` and 1.