Question

Can someone please explain this question to me:

Find values of k for which the expression kx^2 + 3kx + 9 > 0 for all real x.

The answer is 0 < k < 4 but I don't know how to figure that out

Answer 1

the important thing : the expression has to be larger than 0 for all real x (that means for all values, the whole graph of the function has to be above the x-axis).

that is only possible, if the function has NO real zero solution (there are no x-axis interceptions). the only zero solutions are complex/imaginary numbers.

in other words, we need to find all k that would give the function only complex zero solutions.

kx² + 3kx + 9 > 0

the solution to a quadratic equation

ax² + bx + c = 0

is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = (-3k ± sqrt((3k)² - 4×k×9))/(2k) =

= (-3k ± sqrt(9k² - 36k))/2k

to have real x as solution the discriminat (the content of the square root function) must be >= 0.

to create complex numbers as solutions, this discriminant must be therefore < 0 (as a square root of a negative number is a complex number).

9k² - 36k < 0

k² - 4k < 0

to find the interval limits, we can look for the "= 0" solutions, which are then not included in the actual solution (due to the "< 0" condition).

like above

k = (4 ± sqrt((-4)² - 4×1×0))/(2×1) =

= (4 ± sqrt(16))/2 = (4 ± 4)/2

k1 = (4 + 4)/2 = 8/2 = 4

k2 = (4 - 4)/2 = 0

and so we see

0 < k < 4

FYI : we could also continue from

k² - 4k < 0

to find the "= 0" solutions by

k² - 4k = 0

k(k - 4) = 0

k = 0 and k = 4

giving us again

0 < k < 4

Answer 2

0 < k < 4

Explanation:

The given quadratic function will be greater than zero (positive) when its graph is above the x-axis. If the quadratic graph is above the x-axis, it will have no roots since it will not cross the x-axis. To find the values of k for which the quadratic has no roots, use the discriminant.

The discriminant is the part of the quadratic formula underneath the square root symbol: b² - 4ac. It tells us how many real solutions (roots) the quadratic function has.

`\boxed{\begin{minipage}{7.6 cm}\underline{Discriminant}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\when $b^2-4ac > 0 \implies$ two real solutions.\\when $b^2-4ac=0 \implies$ one real solution.\\when $b^2-4ac < 0 \implies$ no real solutions.\\\end{minipage}}`

Given quadratic expression:

`kx^2 + 3kx + 9 > 0`

Therefore:

`a=k,\;\;b=3k,\;\;c=9`

Substitute the values of a, b and c into the discriminant:

`\begin{aligned} \implies b^2-4ac&= (3k)^2-4(k)(9)\\&=9k^2-36k\end{aligned}`

To find the values of k for which the quadratic function has no real roots, set the discriminant to less than 0:

`\begin{aligned} \implies 9k^2-36k& < 0\\9(k^2-4k)& < 0\\k^2-4k& < 0\end{aligned}`

The quickest way to solve this inequality is to consider the graph of f(k) = k² - 4k. As the leading coefficient is positive, the graph is a parabola that opens upwards. To find its roots (the points at which is crosses the x-axis), set it to zero and solve for k:

`\begin{aligned} k^2-4k&=0\\k(k-4)&=0\\\implies k&=0,\;4\end{aligned}`

Therefore, the roots of f(k) = k² - 4k are k = 0 and k = 4.

When k² - 4k < 0, its graph is below the x-axis. The graph is below the x-axis in the interval between its roots: 0 < k < 4.

Therefore, the values of k for which kx² + 3kx + 9 > 0 is 0 < k < 4.