k(x^2-3x+4)+x-3 is always positive if x is a real number. Find the range of k?

Question

k(x^2-3x+4)+x-3 is always positive if x is a real number. Find the range of k?

Answer 1

Hello,



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You are told that for all real values

   𝑘(𝑥² – 3𝑥 + 4) + 𝑥 – 3 > 0



► Obviously, if 𝑘 were equal to 0, we would get:

   0×(𝑥² – 3𝑥 + 4) + 𝑥 – 3 > 0

   𝑥 – 3 > 0

   𝑥 > 3

which is plainly false for all real values of 𝑥.

Thus 𝑘≠0.



► Since 𝑘≠0, we can divide by it. Let 𝑝=1/𝑘, then:

   𝑘(𝑥² – 3𝑥 + 4) + 𝑥 – 3 > 0

   𝑘(𝑥² – 3𝑥 + 4 + 𝑥/𝑘 – 3/𝑘) > 0   ←←← Factoring by 𝑘

   (𝑥² – 3𝑥 + 4 + 𝑝𝑥 – 3𝑝) / 𝑝 > 0   ←←← Since 𝑝=1/𝑘

   [ 𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) ] / 𝑝 >  0



► We want:

  [ 𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) ] / 𝑝 > 0

for all real value 𝑥.



This is only possible if

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝)

have a constant sign, either positive or negative.



Said another way, we want the quadratic equation:

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) = 0

to have NO solution.



► Thus we need to compute the discriminant of the quadratic equation:

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) = 0



This discriminant is:

   𝛥 = (𝑝 – 3)² – 4×1×(4 – 3𝑝) = 𝑝² – 6𝑝 + 9 – 16 + 12𝑝 – 16 = 𝑝² + 6𝑝 – 7 = (𝑝 – 1)(𝑝 + 7)



And we know that if:

   𝛥 < 0, the equation has no solution.

   𝛥 = 0, the equation has an unique solution.

   𝛥 > 0, the equation has two distinct solutions.



► Since we want

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) = 0

to have NO solution, we need then to have:

   𝛥 < 0

   (𝑝 – 1)(𝑝 + 7) < 0



► Since you want a product of two factors to be negative, this means the factors have opposite signs.

So either:

   𝑝 – 1 > 0    and    𝑝 + 7 < 0

   𝑝 > 1          and    𝑝 < -7

which is impossible since a value cannot be at the same time lower that -7 and greater than 1.



Or

   𝑝 – 1 < 0    and    𝑝 + 7 > 0

   𝑝 < 1          and    𝑝 > -7

which can be condensed into:

   -7 < 𝑝 < 1



► Thus, if  -7< 𝑝<1

then

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝)

will have a constant sign.



Since the coefficient of 𝑥² is positive

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝)

will be always positive.



So if -7<𝑝<1,

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) > 0

and there is no value of 𝑝 that would make it negative.



► Coming back to the initial inequation:

   [ 𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) ] / 𝑝 >  0



Since you want a quotient to be positive, this means the top and bottom have the same sign.

Since we know:

   𝑥² + (𝑝 – 3)𝑥 + (4 – 3𝑝) > 0

we conclude we must have

   𝑝>0



Thus we must reduce the solution from -7<𝑝<1 to:

   0 < 𝑝 < 1



► Finally, since we have defined 𝑝=1/𝑘 :

   0 < 𝑝 < 1

   +∞ > 1/𝑝 > 1/1   ←←← Inequality signs needs to be flipped because of inversion

   +∞ > 𝑘 > 1

   𝑘 > 1



► Thus the conclusion:



If  𝑘>1 then we will always have

   𝑘(𝑥² – 3𝑥 + 4) + 𝑥 – 3 > 0

for all real values 𝑥.



Regards,

Dragon.Jade :-)

Answer 2

A previous responder wanted to quibble with the use of the word "range" in the question. But it is obvious that what is being asked is the interval of k's that will cause k(x^2-3x+4) + x - 3 to have a positive value for all real x. Whether we call it a "range" or something else.



Obviously 0 or a negative k will not be satisfactory. When k=1 you have x^2 - 2x + 1 = (x-1)^2 as the value of the expression, and certainly that is never NEGATIVE, but it's 0 when x=1. When k = 2, you have 2x^2 - 5x +5, or 2(x-5/4)^2 - 12.5 + 5, which is negative for small "enough" x. When k = 1.1, you have 1.1x^2 - 3.3x + 4.4 + x - 3 = 1.1x^2 - 2.3x +1.1 = 1.1(x-1)^2 - 0.1x, again can be negative. When k = 0.9, you have 0.9x^2 - 2.7x + 3.6 + x - 3 = 0.9x^2 - 1.7x +0.3, which is negative when x=1.



I think what the teacher or other questioner intended was that k=1 be the only "correct" value...except it's NOT correct, because the given expression can be 0 when k = 1 and x = 1.

Answer 3

The information says the value of the function is always positive, so the range is f(x) > 0

Answer 4

corrected







k(x² - 3x + 4) + x - 3 > 0...........k > 0 ....(A)



(kx² - 3kx + 4k) + x - 3 > 0



kx² - (3k - 1)x + (4k - 3) > 0





D(discriminant) < 0



(3k - 1)² - 4k(4k - 3) < 0



9k² - 6k + 1 - 16 k² + 12k < 0



- 7k² + 6k + 1 < 0



7k² - 6k - 1 > 0



(7k + 1)(k - 1) > 0



k < - 1/7 , 1 < k　............(B)





from 　　(A)(B)　　1 < k

Answer 5

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