Find the greatest integer k such that 3x^2 + (12-3k)x + 15 - 6k is positive for all real values of x?

Question

Need help fast!

the answer should be k=1

Answer 1

Yes I got it.

Given 3x^2 + (12-3k)x + 15 - 6k

First get the 3 common and the equation reduces to
3 { x^2 + (4-k)x + 5 - 2k }

which gives 3 { x^2 + 4x + 5 - kx - 2k }

implies 3 { (x^2 + 4x + 4) + 1 -kx -2k }

Gives 3 { (x+2)^2 + 1 - k(x+2) }

In the above expression, (x+2)^2 + 1 is positive as the range of a perfect square is (0,infinity).

For this expression to be positive, (x+2)^2 + 1 should be grater than k(x+2)

(x+2)^2 + 1 > k(x+2)

gives (x+2) + 1/(x+2) > k --------- (1) (Say)

CONCEPT :
Consider two numbers y and 1/y
Apply AM >= GM to it
It gives y + 1/y >= 2
So blindly follow the following principle.
Therefore the minimum value of the sum of a number and its reciprocal is 2.

Therefore to get the condition the equation (1) changes to the following form.

2 > k (This is the solution for the problem).

The conclusion is, "k is less than 2 but not equal to 2"

Since you want the integer value the next integral value such that the expression will be positive is 1.

Therefore the greatest integer k such that 3x^2 + (12-3k)x + 15 - 6k is positive for all real values of x is 1.

Hope you understand this answer.

Answer 2

So that the function of equations of (x^2) is always positive, this means that the vertex of the graph of this fuction should be above the x-axis.
so we get first of all the x-coordinate of the vertex by the formula
x= -b/2a which is -(12-3k)/(2*3) = (3k-12)/6 = 3(k-4)/6
= (k-4)/2
so we substitute the answer with the unknown x in the equation, to obtain the points with y values greater than 0.

3{ (k-4)/2}^2 + (12-3k) {(k-4)/2} +15 -6k > 0
3 [ { (k-4)/2}^2 + ( 4-k) (k-4)/2 + 5 - 2k ] > 0
so {(k^2) -8k + 16}/4 + { 4k -(k^2) -16 +4k }/2 + 5-2k >0
we multiply by 4:
(k^2) -8k +16 +2*{8k -16 - (k^2) } +20 -8k >0
(k^2) -8k +16 + 16k -32 -2(k^2) +20 -8k >0
-(k^2) +0k +4 >0
(k^2) -4 <0
(k +2) (k -2) <0
k= +2 , -2
so with the number line: -2 so the greatest value of k is 1

Answer 3

A quick mental analysis says the answer could be two but my mind might be forgetting a possible solution. One does sound like a logical answer.

Answer 4

You need to get K by itself. Purplemath.com is a great site for Algebra help. good luck