Question

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

Answer 1

Answer:

C.

Step-by-step 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.

Answer 2

Answer:

Step-by-step 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.