Question

Which statement best describes why there is no real solution to the quadratic equation y = x2 - 6x + 13?

The value of (-6)2 - 4 • 1 • 13 is a perfect square.
The value of (-6)2 - 4 • 1 • 13 is equal to zero.
The value of (-6)2 - 4 • 1 • 13 is negative.
The value of (-6)2 - 4 • 1 • 13 is positive.

Answer 1

Explanation:

Next time, please share the possible answers. Thanks.

Here the coefficients are a = 1, b = -6 and c = 13. Let's calculate the determinant b^2 - 4ac: d = (-6)^2 - 4(1)(13) ) = 36 -52 = -16.

Because the determinant is negative, this quadratic has only complex roots. The third answer applies here.

Answer 2

C.

Explanation:

`y=x^2-6x+13` when compared to
`y=ax^2+bx+c` tells us:

`a=1`

`b=-6`

`c=13`.

The discriminant,
`b^2-4ac` tells how many real solutions we will have.

If
`b^2-4ac` is zero then you have one real solution.

If
`b^2-4ac` is positive then you have two real solutions.

If
`b^2-4ac` is negative then you have no real solutions.

`b^2-4ac`

`(-6)^2-4(1)(13)`

`36-4(13)`

`36-52`

`-16`

Our discriminant is negative, so we have no real solutions.

The answer you are looking for is the one that says your discriminant is negative which is C.