## gjhfdfg 2 years ago Finding x intercepts for quadratic functions

1. gjhfdfg

I need help on finding the x intercept on this quadratic function, |dw:1360627184057:dw|

2. Marx

3. gjhfdfg

I do now I think,

4. gjhfdfg

Hold on,

5. gjhfdfg

Ok, so I solve it like this? |dw:1360627636615:dw|

6. Marx

almost, a= 6 not 6x^2

7. Marx

that way you have a number :D

8. gjhfdfg

What happens to the x^2?

9. Marx

well: a,b and c are the coefficients so there's no need for x's. check the formula again. so you should have two solutions.

10. gjhfdfg

Them 2 solutions got me lost..

11. jazy

|dw:1360628346901:dw|

12. gjhfdfg

I got that so far,

13. Marx

it goes like this: $x_{1,2} =\frac{-12\pm \sqrt{12^2-4\cdot 6 \cdot 5}}{6\cdot 2} = -1\pm \frac{\sqrt{24}}{12}=-1\pm\frac{2\sqrt{6}}{12}=-1\pm\frac{1}{\sqrt6}$ so we have : $x_1 = 1+\frac{1}{\sqrt{6}}$ and x_2 = 1-\frac{1}{\sqrt{6}}

14. Marx

i mean $x_2= 1-\frac{1}{\sqrt{6}}$

15. Marx

ahem, yeah with -1 ... :)

16. gjhfdfg

How did you get the$\sqrt{24}$ ?

17. Marx

$12^2 - 4\cdot 6 \cdot 5 = 144 - 120 = 24$

18. gjhfdfg

Ah, ok I was thinking 12*12 = 120 instead of 144. My bad,

19. gjhfdfg

But then how did $2\sqrt{6}$ come in?

20. gjhfdfg

The square roots throw me off...

21. Marx

$\sqrt{24}=\sqrt{4\cdot 6} =\sqrt{4} \cdot \sqrt{6} = 2\sqrt{6}$

22. gjhfdfg

Ah ok got it thanks.!

23. Marx

you welcome :)

24. gjhfdfg

Do we do anything else to it?

25. Marx

what do you mean?

26. gjhfdfg

Is there anything else I should do to the answer(s)? None of them match my answer choices

27. Marx

28. gjhfdfg

29. Marx

so it's the second one: $\frac{-6\pm \sqrt{6}}{6}$ it's basically what I told you but in a different form

30. gjhfdfg

Ah ok thank you

31. Marx

$\frac{-6\pm\sqrt{6}}{6} = -1 \pm \frac{\sqrt{6}}{6} = -1\pm \frac{1}{\sqrt{6}}$

32. gjhfdfg

Got it.!