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anonymous
 one year ago
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
anonymous
 one year ago
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

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0focus on "b^2 4ac " as you need to take its square root.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, I'm not sure how to solve it. Can you teach me? @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0what is b and what is c in your original quadratic?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0this is the quadratic you posted: \(\large x^2  10x = 13 \) and this is the additional formula you also posted: \(\large x = \frac{b \ \pm \ \sqrt{ b^2 4ac} }{ 2a}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0maybe 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I'm unsure on how to solve this question...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0that's my question....you are trying to find the value of x where the quadratic = 0, is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, the question is asking to find complex solutions... Which I dont really understand what it means by that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I guess it's asking to solve also.. So yeah. solving for x

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0take your original equation \(x^2  10x = 13 \) and rewrite 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.....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0clearly if \(b^2  4ac \lt 0\) you might have an issue (ie complex roots) but run with it first....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, I'm sorta confused on how to plug it in and solve it... Can you help me on that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 you there?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0compare \(x^2−10x+13=0\) to \(ax^2+bx+c=0\) so \(a = 1, b = 10, c = 13\) plug these into \(\huge x = \frac{b \pm \sqrt{b^2  4ac}}{2a}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438365369887:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, I regret not paying enough attention in math class... How do I solve this? Sorry to take up more of your time by asking this. I never got anything out of quadratics... *sigh*

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0cool, but its \(b \pm\) and (10) = 10 but now get out your calculator/spreadsheet/whatever and calculate the 2 values of x. it is \(\pm\) so there will be 2 values. these are your roots

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(\huge \frac{(10) \pm \sqrt{(10)^2  4(1)(13)}}{2(1)} \implies \) \(\huge \frac{10 \pm \sqrt{100  52}}{2} = 1.54, \ 8.46 \)
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