Quadratic Equations

bonkysleuth SG

Find the range of values of c for which 3x^2 - 6x +c > 4 for all real values of x.

so for me,

I wrote that

D>0

After manipulating with the equation i got

84 - 12 c > 0

12 (7-c) >0

so c <7

However my mark sheet gives the answer as c>7. To check whether my approach is correct, i looked at the TYS (whcih contains some similar questions) and yes, i got thsoe questions in the TYS correct, but why is it wrong for this question? can someone clarify? and may i know, for such questiosn where they do not tell you if the roots cut 2 points/1point/no point, is the discriminant always that of the sign given in the equation?

and are such questions different from "ax^2 + bx + c is always positive/negative where D is always <0"?

wee_ws SG

Hi,

Before we conclude about the nature of D, we need to rearrange the inequality to

3x^2 - 6x + c - 4 > 0.

Thus we want to look at the case for which D < 0. This question belongs to the category of "ax^2 + bx + c is always positive/negative for any real x".

Thanks!

Cheers,
Wen Shih