The Student Room Group
a=1a=1

b=2kb=-2k

c=k+6c=k+6
Reply 2
a = 1
b = -2k
c = k + 6

That help? :smile:
If there are no real roots b24ac. b^2 - 4ac . (is less than zero)

Sub a,b and c then draw a graph if it will help you find the range of values ( b^2 - 4ac is less than 0 )

Spoiler

Reply 4
Graph? Why?
Mr M
a=1a=1

b=2kb=-2k

c=k+6c=k+6


ahh okay lol
Thanks + rep for you :smile:
No need for a graph if you can do it all alegebraically
Reply 7
Your expression you wrote is wrong.

It's b^2 - 4ac < 0



K must be greater than -3 and less than 2.
Guinny
Graph? Why?


Drawing a graph hepls you to see the solutions toa qaudratic inequality. In my opinion.
Reply 9
DaveJ
Your expression you wrote is wrong.

It's b^2 - 4ac < 0


b24ac<0b2<4acb^2 - 4ac < 0 \Leftrightarrow b^2 < 4ac
so [br]2k<4k+24[br]-2k < 4k +24
[br]6k<24[br]-6k < 24
[br]k>4[br][br]k > -4[br]

Is that right? when you divide by a negative the more than sign changes the other way.
Thanks again
Reply 11
Rhys.
b24ac<0b2<4acb^2 - 4ac < 0 \Leftrightarrow b^2 < 4ac



Oh... fair enough then. Never seen it written like that before.


OP: Find the discriminant, then factorise it. Sketch the graph, notice that the y values are less than zero between -3 and 2, and that's your answer. That's if I haven't made a mistake, and I probably will have made one somewhere...
Reply 12
Spongebob*No*Pants
so [br]2k<4k+24[br]-2k < 4k +24
[br]6k<24[br]-6k < 24
[br]k>4[br][br]k > -4[br]

Is that right? when you divide by a negative the more than sign changes the other way.
Thanks again


Yes - that's right. easy way to check, sub -4 into the discriminant and see if it = 0, then you can see that if K < -4 , than b^2 is smaller than 4ac
Reply 13
Spongebob*No*Pants
so [br]2k<4k+24[br]-2k < 4k +24
[br]6k<24[br]-6k < 24
[br]k>4[br][br]k > -4[br]

Is that right? when you divide by a negative the more than sign changes the other way.
Thanks again

...
(2k)2<4k+24(-2k)^2 < 4k +24

I would shift it all to the LHS, simplify and complete the square which would give the possible range for k

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