Question

using the quadractic formula solve 3x^2-17=0

Answer 1

Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 2

can u find a,b,c ?

Answer 3

no

Answer 4

a=3,b=0,c=-17
now just put these in the formula i gave ....

Answer 5

oh ok i get it now

Answer 6

good, so what 2 values of x u got ?

Answer 7

i got 2.38 and -2.38

Answer 8

that is correct!
good work :)

Answer 9

how about this equation 2y^2=-5y

Answer 10

here a=2, b= 5,c=0
do u get how ?

Answer 11

so would the answers be 2.28 and .22

Answer 12

in the back of the book they have -5/2,0

Answer 13

do u need to use formula ? because there is simpler way for 2y^2=-5y

Answer 14

in the book it says that you do

Answer 15

so did u get how a=2, b= 5,c=0

Answer 16

then u just put these in formula

Answer 17

i did put them in the formula but how do you get -5/2 and 0?

Answer 18

ok, what u got b^2-4ac as ?

Answer 19

5^2-4*2*0 = ?

Answer 20

yes i have that

Answer 21

so what numerator u got ?

Answer 22

the denominator =2a=4

Answer 23

\(\large -b+\sqrt{b^2-4ac}=-5+\sqrt{25-0}=-5+5=0\)
did u get this ?

Answer 24

so one of the root = 0

Answer 25

\(\large -b-\sqrt{b^2-4ac}=-5-\sqrt{25-0}=-5-5=-10\)
and with denominator =4, u get other root as -10/1 = -2/5
got this ?

Answer 26

*-10/4=-5/2

Answer 27

ok

Answer 28

good, any more doubts ?

Answer 29

no that is it

Answer 30

thanks

Answer 31

welcome ^_^