Solve Inequality Problem for f(x)=x^2+kx+k, Determine k

[Sisyphus]

for f(x)=x^2+kx+k, determine all values of k such that f(x)>0

could someone please help me get started here? I am reviewing for a high school math contest and it's been quite some time since I've learned about inequalities. I've managed to solve some rudimentary problems, but this one really has me stumped.

 

[greytomato]

simple enough. this equation represents a curve called parabula.
try drawing this curve on XY surface.
try with different k values such as: k=0, k=1, k=-1... and see how this changes the curve.

*please note that k=0 doesn't match your question because if does get the value of 0 and you're required to find f(x)>0.

if you need more help, let me know how it goes...

 

[benorin]

Try completing the square: x^2+kx+k=\left( x+\frac{k}{2}\right) ^2+k-\frac{k^2}{4}>0 for what values of k?

 

[greytomato]

since this is a parabula, you can find its roots (the points where f(x)=0). there are always 2 roots.
roots equations: (-B +- sqrt(B^2 - 4AC)) / 2A
A is the coefficient of X^2
B is the coefficient of X^1
C is the coefficient of x^0

the parabula has 3 different possible conditions:
#1: intersecting the X axis twice.
#2: intersecting the X axis once.
#3: not intersecting the X axis at all.

think which one you want it to be, and use the roots equation to calculate accordingly

 

[benorin]

For a fixed value of k, f(x) is an upwards opening parabola whose vertex occurs at x=-\frac{k}{2}, we require that all points on the parabola be above the x-axis and hence require that the lowest point on the parabola be above the x-axis, that is the y-coordinate of the vertex must be >0, i.e. we require f\left( -\frac{k}{2}\right) >0 .

 

[greytomato]

@benorin: I'm not sure your suggestion leads to the full solution. i might be wrong though...

 

[daveb]

Benorin's solution works completely because it's the same as yours (using the discriminant).

 

[Sisyphus]

Thanks a lot, benorin and greytomato!

Visualising how the parabola whose vertex is above y=0 was the connection that I had needed to make on my own. :P From there, I just completed the square, and solved for the values of k that would keep the vertex above y=0. I probably wouldn't have thought to use discriminants that way though, so it's good that I saw an alternative way of solving the problem.

Another problem came up for me today, and I think it's probably better to post it in this thread instead of clogging the forums up with a new thread:

Find the minimum value of f(x)=3^(x^2+4x)

For this question, I figured f(x)>0, since a number raised to a power cannot yield a negative number. so, I have

0<3^(x^2+4x)

written down, but once again, I find myself stuck =\

 

[VietDao29]

Thanks a lot, benorin and greytomato!

Visualising how the parabola whose vertex is above y=0 was the connection that I had needed to make on my own. :P From there, I just completed the square, and solved for the values of k that would keep the vertex above y=0. I probably wouldn't have thought to use discriminants that way though, so it's good that I saw an alternative way of solving the problem.

Another problem came up for me today, and I think it's probably better to post it in this thread instead of clogging the forums up with a new thread:

Find the minimum value of f(x)=3^(x^2+4x)

For this question, I figured f(x)>0, since a number raised to a power cannot yield a negative number. so, I have

0<3^(x^2+4x)

written down, but once again, I find myself stuck =\

One big hint for you is that the function f(x) := 3^x is a strictly increasing function, that is for every x₁ < x₂, you'll always have:
f(x₁) < f(x₂).
Now to find the minimum value of 3 ^ {x ^ 2 + 4x}, you must find a minimum value of: x² + 4x, which can be done by completing the squares.
Can you go from here? :)

 

[Sisyphus]

yes! thank you very much!