# How to find the possible ranges of an equation

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#1
I have the following question:

Find the possible values of k if the equation g(x) = 0 is to have two distinct real roots where g(x) is given by g(x) = 3kx^2 + kx + 2

I have correctly worked out the answer (I also know it is > 0 because it has two real roots):

a = 3k
b = k
c = 2

b^2 - 24k > 0
k(k - 24) > 0
so either k = 0 or k = 24

In the answers I get k < 0 or k > 24
How do I know that k < 0 or k > 24
0
7 months ago
#2
(Original post by Simon33355)
b^2 - 24k > 0
k(k - 24) > 0
so either k = 0 or k = 24
It doesn't make sense for you to go from k(k-24)>0 to "k=0 or k=24".

It only make sense if you had k(k-24)=0.

Since this is an inequality, your answer should involve inequalities for k.

It's quite simple to deduce the answer however. Are you able to sketch the quadratic k(k-24) ?? I sure hope so because it should be straightforward, and with that sketch, note that k(k-24) > 0 asks you "for what values of k is the quadratic above the horizontal axis?"

So you look at your sketch, and hopefully you can see where that final answer comes from.
1
7 months ago
#3
You can’t find the square root of a negative number to get a real root and square rooting 0 only gives one root. When you have two distinct real roots your discriminant must be greater than 0 so that you have a +- answer allowing you to have two real roots (two points of intersection on a cartesian graph). Thus to find the values of k that satisfy this you let the discriminant be greater than 0. Substituting for a b and c allows you to find the range of values for k thay satisfy this condition. You know that these are correct (assuming no arithmetic error) because you created the equation on the basis of two distinct real roots using basic facts such as the discriminant needs to be greater than 0.
1
#4
(Original post by RDKGames)
It doesn't make sense for you to go from k(k-24)>0 to "k=0 or k=24".

It only make sense if you had k(k-24)=0.

Since this is an inequality, your answer should involve inequalities for k.

It's quite simple to deduce the answer however. Are you able to sketch the quadratic k(k-24) ?? I sure hope so because it should be straightforward, and with that sketch, note that k(k-24) > 0 asks you "for what values of k is the quadratic above the horizontal axis?"

So you look at your sketch, and hopefully you can see where that final answer comes from.
Thank you, that makes sense. Since the shape of the graph is curved, and g(x) must be more than 0, then the curve k > 24 and k < 0 would be above 0 on the axis.
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