Which statement best describes why there is no real solution to the quadratic equation y = x^2 - 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.

Question

Which statement best describes why there is no real solution to the quadratic equation y = x^2 - 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.

mathstudent55 · Answer

Are you familiar with the quadratic formula?

josedavid · Answer

no

josedavid · Answer

so is a

mathstudent55 · Answer

The solution to the quadratic equation

$ax^2 + bx + c = 0$

is

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

This last equation is the quadratic formula.

mathstudent55 · Answer

Have you seen it before?

mathstudent55 · Answer

The part of the quadratic formula that is inside the square root in the numerator, 
$b^2 - 4ac$, is called the determinant.
It lets you determine the type of solutions the quadratic equation has.

If the determinant is greater than zero, there are 2 different real solutions.
If the determinant equals zero, there is one real solution.
If the determinant is negative, there are two complex (imaginary) solutions.

mathstudent55 · Answer

Since you have $y = x^2 - 6x + 13$, you have $a = 1$, $b = -6$, and $c = 13$.
Use these values in $b^2 - 4ac$ to evaluate the determinant.
Then use the previous response to determine the nature of the solutions of the quadratic equation.

josedavid · Answer

so is d

phi · Answer

no real solution

How do you get "no real solution"  i.e. how do you get a *complex* or *imaginary* number?
answer: by taking the square root of a negative number.

josedavid · Answer

ummm

phi · Answer

math posted the formula and it has a square root.  what is inside the square root?
in letters:  b^2 -4 a c
with numbers:   (-6)^2 - 4*1*13

josedavid · Answer

is d

josedavid · Answer

hello

josedavid · Answer

@phi

phi · Answer

you need a *negative* number under the square root to get a non-real root.

mathstudent55 · Answer

I wrote above that you need to evaluate the expression
$b^2 - 4ac$
with the values of your trinomial.

$(-6)^2 - 4(1)(13) $

Now you follow the order of operations.
Do the exponent first.

$= 36 - 4(1)(13) $

Now do the multiplication.

$= 36 - 52 $

$= - 16$

Now compare the value above with the choices.

mathmale · Answer

josedavid:  I don't see any work on your part; all I see is your repeated questions, "so is a," "so is d," and so on.  Responding in this way is not going to help you learn the quadratic formula and its applications.

Four possible results are given to you.  What do they mean?  If you'd look up "quadratic formula," you'd likely find interpretations of these outcomes.