## anonymous 5 years ago I have a question for using quadratic equation again. I have 3^2-5x-7=0 Can any one help me?

1. anonymous

Do you mean 3x^2?

2. anonymous

is nt in the form of quadartic

3. anonymous

yes I meant 3x^2-5x-7=0

4. anonymous

(3x+7)(x-1)

5. anonymous

no sorry

6. anonymous

x[1] = 1/6 (5-sqrt(109)) x[2] = 1/6 (5+sqrt(109))

7. anonymous

OK doesn't factorise so use formula: x =( 5 +/- sqrt(25 - 4 times 3 times -7))/3 I get x = (5 +/- sqrt109)/3

8. anonymous

?

9. anonymous

Do you know the quadratic formula? The solution of ax^2 + bx + c = 0 is $x = -b +/- \sqrt{(b^2 - 4ac}) \over 2a$

10. anonymous

yes I know that

11. anonymous

@ suzi20 would it not be this x=-5+sqrt109/6, x=-5-sqrt109/6

12. anonymous

OK so look at suzi20's correct solution and see if you can get the same answer by substituting for a, b and c in the formula, where a = 3, b = -5 and c = -7

13. anonymous

leahp x=-5+sqrt109/6, x=-5-sqrt109/6 is a correct answer

14. anonymous

thank you, again

15. anonymous

I have one more question does the letter i is not also use for the symbol +- sign together