## dddan Group Title if a=2 ; b=-5; c=5 how would you solve the quadratic equation? 2 years ago 2 years ago

1. hartnn

then the two roots of x are: $$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

2. wio

Use the quadratic foumula $x_1 =\frac{ -b + \sqrt{b^2-4ac}} {2a}$ $x_2 =\frac{ -b - \sqrt{b^2-4ac}} {2a}$

3. dddan

ya but i got 4+- the square root of -24 / 4 does this make sense?

4. hartnn

your b=-5, how did u get 4+/- ... ?

5. dddan

sry the b=-4 and the c=5

6. wio

You have imaginary roots. Is that what you're worried about?

7. dddan

yup

8. wio

Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients

9. hartnn

u actually have imaginary roots for this..

10. dddan

alright can u tell me where to take it from the imaginary roots?

11. hartnn

how did u get square root of -24 ?

12. dddan

because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?

13. hartnn

lol! b=-5 not -4

14. dddan

no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5

15. hartnn

ok, $$\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}$$

16. dddan

alright thank you very much!