cameronkinney518
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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.

Respuesta :

Answer:

C.

Step-by-step explanation:

[tex]y=x^2-6x+13[/tex] when compared to [tex]y=ax^2+bx+c[/tex] tells us:

[tex]a=1[/tex]

[tex]b=-6[/tex]

[tex]c=13[/tex].

The discriminant, [tex]b^2-4ac[/tex] tells how many real solutions we will have.

If [tex]b^2-4ac[/tex] is zero then you have one real solution.

If [tex]b^2-4ac[/tex] is positive then you have two real solutions.

If [tex]b^2-4ac[/tex] is negative then you have no real solutions.

[tex]b^2-4ac[/tex]

[tex](-6)^2-4(1)(13)[/tex]

[tex]36-4(13)[/tex]

[tex]36-52[/tex]

[tex]-16[/tex]

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.

altavistard

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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