Question

What indicators predict that a quadratic funtion will have a complex solution?
[Solve and show all work]
x^2 - 10x = -13

x = -b +- squareroot b^2 -4ac OVER 2a

Answer 1

focus on "b^2 -4ac " as you need to take its square root.....

Answer 2

Well, I'm not sure how to solve it. Can you teach me? @IrishBoy123

Answer 3

what is b and what is c in your original quadratic?

Answer 4

this is the quadratic you posted:

\(\large x^2 - 10x = -13 \)

and this is the additional formula you also posted:

\(\large x = \frac{-b \ \pm \ \sqrt{ b^2 -4ac} }{ 2a}\)

Answer 5

maybe it would be better to start by asking you what you are trying to do to the quadratic? i know but do you also know?

Answer 6

No, I'm unsure on how to solve this question...

Answer 7

that's my question....you are trying to find the value of x where the quadratic = 0, is that right?

Answer 8

Well, the question is asking to find complex solutions... Which I dont really understand what it means by that.

Answer 9

and I guess it's asking to solve also.. So yeah. solving for x

Answer 10

take your original equation
\(x^2 - 10x = -13 \) and re-write it as
\(x^2 - 10x +13 =0\) , yep?!!

then compare it to this
\(a x^2 + b x + c = 0\)
and then apply that second equation you posted, the one with "\(b^2 - 4 ac\)" in it [i set it out again above]

that second equation gives you the roots (x intercepts) of the quadratic, ie x values for when quadratic = 0

try it and see where you go.....

Answer 11

clearly if \(b^2 - 4ac \lt 0\) you might have an issue (ie complex roots) but run with it first....

Answer 12

Well, I'm sorta confused on how to plug it in and solve it... Can you help me on that?

Answer 13

@IrishBoy123 you there?

Answer 14

compare
\(x^2−10x+13=0\)
to
\(ax^2+bx+c=0\)

so
\(a = 1, b = -10, c = 13\)

plug these into

\(\huge x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Answer 15

what do you get?