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Oksanka [162]
3 years ago
8

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

Mathematics
2 answers:
NeX [460]3 years ago
7 0

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.

seropon [69]3 years ago
4 0

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.

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