## 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

1. IrishBoy123

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

2. anonymous

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

3. IrishBoy123

4. IrishBoy123

this is the quadratic you posted: $$\large x^2 - 10x = -13$$ and this is the additional formula you also posted: $$\large x = \frac{-b \ \pm \ \sqrt{ b^2 -4ac} }{ 2a}$$

5. IrishBoy123

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?

6. anonymous

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

7. IrishBoy123

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

8. anonymous

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

9. anonymous

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

10. IrishBoy123

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.....

11. IrishBoy123

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

12. anonymous

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

13. anonymous

@IrishBoy123 you there?

14. IrishBoy123

compare $$x^2−10x+13=0$$ to $$ax^2+bx+c=0$$ so $$a = 1, b = -10, c = 13$$ plug these into $$\huge x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

15. IrishBoy123

what do you get?

16. anonymous

|dw:1438365369887:dw|

17. anonymous

okay, 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*

18. IrishBoy123

cool, 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

19. anonymous

Okay.

20. IrishBoy123

$$\huge \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(13)}}{2(1)} \implies$$ $$\huge \frac{10 \pm \sqrt{100 - 52}}{2} = 1.54, \ 8.46$$

21. IrishBoy123

yep?

22. anonymous

Yeah. Thank you.