Question

4.Given that y=2x^2+3x, find the range of values of x for which

Answer 1

\[y>9\]

Answer 2

@hartnn @ganeshie8 @mathmath333

Answer 3

did you first try yourself ?

Answer 4

its the same as last one, give it a try, show your steps

Answer 5

i did try @hartnn

Answer 6

okay

Answer 7

\[9=2x^2+3x\]
\[-2x^2-3x+9=0\]
\[2x^2+3x-9=0\]

Answer 8

good
how will you factorize that ?

Answer 9

nopes,
find 2 numbers whose sum is +3 and product is -18

Answer 10

okay :)

Answer 11

\[-3\times6=\]

Answer 12

-18

Answer 13

\[-3+6=3\]

Answer 14

@hartnn

Answer 15

correct
so now we can factorize
\(2x^2+3x-9 = 2x^2 +6x-3x-9 \\ = 2x (x+3)-3(x+3)\)

can u continue ?

Answer 16

factor out x+3

Answer 17

x=-3 and x=3/2

Answer 18

yeah,
but keep it in factored form

Answer 19

(x+3)(2x-3)

Answer 20

okay

Answer 21

\((x+3)(2x-3) > 0 \)
how will u go ahead ?

Answer 22

\[x >-3~and~x >\frac{ 3 }{ 2 }\]

Answer 23

correct :)

Answer 24

thnx @hartnn