Question

Find the set value of k such that 3x^2-6x+k>0 for all real value of x

Answer 1

Ishaan to rescue!

Answer 2

pls help...

Answer 3

I am having some OS problems


\[b^2 -4ac>0\]Use this

Answer 4

so
(-6)^2-4(3)(k)>0
k<3

but the ans is K>3...

Answer 5

yes that is right!

Answer 6

yes it is!

Answer 7

\[36-12k>0\]
\[36>12k\]
\[3>k\] got it

Answer 8

When you move the 36 to the right side, the sign switches. k>3.

Answer 9

The sign only switches when you multiply or divide the inequality by a negative number.

Answer 10

i don't move in math. i add subtract multiply and divide.
\[36-12k>0\] add 12k
\[36>12k\] divide by 12
\[3>k\]

Answer 11

Neon!

Answer 12

@neoh you are right for sure. answer is
\[3>k\] or you could write
\[k<3\]

Answer 13

i know the rules... but the real answer is k>3...

Answer 14

ooooooooooooooooooooooooooh i see

Answer 15

we were all entirely wrong

Answer 16

you want this to be always positive, i think we read it to have "real zeros"
if it is always positive it has NO real zeros in which case
\[b^2-4ac<0\]

Answer 17

of course you have already solved this, but backwards

Answer 18

the exact answer in the book is {k:k>3}... but i need to know the step....

Answer 19

hope it is clear:
always positive
sits above the x -axis (never crosses or touches)
has no real zeros
therefore the discriminant is NEGATIVE

Answer 20

if not clear let us know. but i repeat, if
\[3x^2-6x+k>0\] for all x, that means
\[3x^2-6x+k\] is never zero, which means
\[(-6)^2-4\times 3\times k<0\]
\[36-12k<0\]
\[k>3\]

Answer 21

ok... if not clear then i tell u i try to think 1st...thx for the answer...

Answer 22

i think i ald get it... thx for d ans....

Answer 23

yw