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16)
#kx^2-2kx=5#
#kx^2-2kx-5=0#
The standard form of the quadratic equation is:
#ax^2+bx+c=0#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#b^2-4ac# is called the discriminant of the equation.
If the discriminant has a positive value there are two solutions for #x#.
If the discriminant is negative there are no real solutions for #x#. The solutions will be complex numbers.
If the discriminant is equal to zero there is only one solution for #x#.
In this problem:
#a=k#, #b=-2k#, and #c=-5#. Let's compute the discriminant:
#b^2-4ac=(-2k)^2-4(k)(-5)=4k^2+20k#
Let's set the discriminant equal to zero and find its roots:
#4k^2+20k=0#
#4k(k+5)=0#
#k=0, -5#
#k=0# is not possible because it gives us #x=0/0#.
For #k=-5#, the discriminant is equal to zero and #x# has only one solution. #x=-b/(2a)=(2k)/(2k)=(-10)/(-10)=1#
For #4k^2+20k<0#, there are no real solutions for #x#. This happens when #k<-5#
For #4k^2+20k>0#, there are two solutions for #x#. This happens when #k>(-5)#
17)
#a=k-3#, #b=2k#, and #c=k+2#
Discriminant is #b^2-4ac=(2k)^2-4(k-3)(k+2)=#
#4k^2-4(k^2-k-6)=4k^2-4k^2+4k+24=4k+24#
To have two solutions for #x#:
#4k+24>0#, that is #k>(-6)#
To have one solution for #x#:
#4k+24=0#, that is #k=-6#
More Detailed Explanation:
'__________'
Remember that your function is #y# as a function of #x#. Ordinarily, you would not even have #k# in the picture. You are just trying to find values of #k# that will give you either one or two solutions for #x#. You should not be trying to graph #4k+24#.
Let's try two values for #k#, one less than #-6# and one greater than #-6#. For less than #-6#, we will try #k=-7#:
#y=(k-3)x^2+2kx+(k+2)=0#
#(-7-3)x^2+2(-7)x+(-7+2)=0#
#-10x^2-14x-5=0#
#-(10x^2+14x+5)=0#
#10x^2+14x+5=0#
#x=(-14+-sqrt(14^2-4(10)(5)))/(2(10))=(-14+-sqrt(-4))/20=#
#(-14+-sqrt(4(-1)))/20=(-14+-2sqrt(-1))/20=(2(-7+-sqrt(-1)))/20=#
#(-7+-sqrt(-1))/10#
From the domain of complex numbers, we know:
#i^2=-1#, and #i=sqrt(-1)#
Therefore,
#x=-7/10+1/10i#, and #x=-7/10-1/10i#
Both are complex numbers and you do not have any real solutions for #x#. Now, let's see what the graph looks like:
graph{10x^2+14x+5 [-2.563, 2.437, -0.42, 2.08]}
As, you can see, the graph is a parabola opening in the positive direction (up) but does not cross the #x#-axis. As such, there are no solutions for #x#.
Now, let's try a value greater than #-6# for #k#. We will try #k=-5#:
#(-5-3)x^2+2(-5)x+(-5+2)=0#
#8x^2+10x+3=0#
#x=-1/2#, and #x=-3/4#.
Both are real numbers and you have two solutions for #x# as we can see in the graph below:
graph{8x^2+10x+3 [-1.682, 0.818, -0.24, 1.01]}
The graph is crossing the #x#-axis at two points whose #x# values are the two answers we found.
If we try #k=-6# we will have only one solution for #x#:
#(-6-3)x^2+2(-6)x+(-6+2)=0#
#9x^2+12x+4=0#
#x=(-12+-sqrt(12^2-4(9)(4)))/(2(9))=(-12+-sqrt(144-144))/18=(-12+-0)/18=-12/18=-2/3#
Now, let's see the graph:
graph{9x^2+12x+4 [-1.3145, 0.333, -0.136, 0.688]}
As you can see, the graph only touches the #x#-axis at the value of #x# we found. Therefore, you only have one solution.
As for your question of trying #k# values in the equation, if you do it will either verify you have the right answer or let you know you may have made a mistake. It is not a bad idea.
