Hi! Please, help to solve the inequality :
1) -х²+3x-5>0 ;
2) 3х²-4x+8≥0 .
Tnk U!
@Dan It would best if this were moved to a new thread.
@AnnaPorter. Please begin a new thread when posting a new question.
1) Now that has no real roots bec, $(3)^2-4(-1)(-5)<0$ SEE HERE
BUT $x=1$ we have $-(1)^2+3(1)-5<0 so there is non solution. WHY?
2) The discriminate here is $(-4)^2-4(3)(8)<0 $ again there are no real roots SEE HERE
but this for $x=1$ we have 3(1)^2-4(1)+8>0$ so that means it is true everywhere. WHY?
Another way of looking at these:
Completing the square
$\displaystyle -x^2+ 3x- 5= -(x^2- 3x+ (9/4- 9/4))- 5= -(x^2- 3x+ 9/4)+ 9/4- 5$
$\displaystyle -(x- 3/2)^2+ 9/4- 20/4= -(x- 3/2)^2- 11/4$
We can think of this as -11/4 minus a square so it is never larger than -11/4 so never larger than 0.
$\displaystyle 3x^2- 4x+ 8= 3(x^2- (4/3)x+ (4/9- 4/9))+ 8= 3(x^2- (4/3)x+ 4/9)- 4/3+ 8$
$\displaystyle 3(x+ 2/3)^2- 4/3+ 24/3= 3(x+2/3)^2+ 5$
That is 5 plus a square to is never less than 5 and so never less than 0.