What Values of K Make the Intersection of Two Graphs Positive?

[TyErd]

the graph of y=kx+3 intersects the graph of y=x^2 +8x at two distinct points for
'k 'equals what?

to be honest I do not know where to start

 

[zeion]

It's a parabola and a line.

 

[TyErd]

yeah...but how do you find k

 

[Mark44]

At any point of intersection of the two curves, the coordinates x and y will satisfy both equations.

You have y=kx+3 and y = x^2 + 8x. Set the right sides of these equations equal and solve for x.

 

[TyErd]

how can you solve for x when you don't know k

 

[Bohrok]

You're not finding an exact value for x, just getting it by itself and leaving the equation in terms of k.

 

[Mentallic]

x will be in terms of k.
Once you set the quadratic in the form $ax^2+bx+c$ with a,b,c any number or variable in terms of k, use the quadratic formula to solve for x.

Now, once you have x=... what do you need in order to have 2 distinct points? It is the same thing as thinking of what you would need to have 2 distinct roots to a quadratic equation.

 

[TyErd]

Can anyone actually show me their working out so I can gain a more coherent understanding?

 

[HallsofIvy]

Better if you show us your working. You learn math by doing, not by watching!

As Mark44 told you to start with, set y= kx+ 3= x^2+ 8x. Write that in the form ax^2+ bx+ c= 0 and use the quadratic formula. Yes, there will be a k in the formula.

You don't have to actually solve the equation to answer this question. Since the question asks about two different points of intersection, you want to decide what k must be so that equation has two distince solutions. A quadratic equation can have 0, 1, or 2 solutions depending on the "discriminant", b^2- 4ac.

 

[TyErd]

https://www.physicsforums.com/showthread.php?t=356820

 

[TyErd]

so once rearranged it will be x^2+(8x-kx)-3=0
Use the quadratic formula and it is: (-8+k +/- sqrt k^2-16k+76 ) / 2
am i correct up to that point?

 

[HallsofIvy]

Yes, what you have is correct.

And, as I said, you really only need the discriminant: (8- k)^2+ 12= k^2-16k+ 76. For what values of k is that positive?

 

[Mark44]

so once rearranged it will be x^2+(8x-kx)-3=0
Use the quadratic formula and it is: (-8+k +/- sqrt k^2-16k+76 ) / 2
am i correct up to that point?

You are solving for x, so you really should have x on one side of an equation.

x = [(-8 + k) +/- sqrt(k^2 - 16k + 76)]/2

 

[TyErd]

Sorry HallsofIvy I am having trouble understanding what you meant by "For what values of k is that positive".

the discriminant is k^2-16k+76, what do you do with that

 

[Mark44]

You have k^2 - 16k + 76 as the discriminant. For what values of k is the discriminant > 0? That's what HallsOfIvy asked. Do you know how to solve a quadratic inequality?

 

[TyErd]

umm..unfortunately i don't

 

[Mark44]

Do you know how to solve a quadratic equation?

 

[TyErd]

By using the quadratic formula?

 

[Mark44]

Or however.

 

[TyErd]

okay so i have k^2-16k+76, if i use the quadratic formula it is (16+/-sqrt (-16)^2 - 4*1*76 )/ 2, but this gives an error on my calculator

 

[Mentallic]

Just like when you solve any quadratic, when you get the error (which means the discriminant is negative, and you can't take the square root of a negative), how does this describe the properties of the parabola?

For e.g. Take a look at the graph of $y=x^2-1$
and solve x in $x^2-1=0$

Now take a look at $y=x^2+1$
and solve x in $x^2+1=0$

can you see what is happening and understand why you get an error on the second one?
When you're solving for x, you're basically saying where does y=0.

 

[TyErd]

yeah okay i get that, so how am i suppose to get value/s for k

 

[Mentallic]

Well, for what values of x is $x^2+1>0$ ?

In a similar fashion, the discriminant was $(k-8)^2+12$. For what values of k is $(k-8)^2+12>0$. And the answer to this gives you the values of k where the quadratic and the line has 2 distinct roots.

 

[TyErd]

ohk, so (k-8)^2+12>0
(k-8)^2>-12

then what would do since you cannot square root a negative

 

[Mentallic]

No, no. I'm just asking you to use logic here.
What are the possible values of $x^2$ for all x? In other words, what is the range of $x^2$ for all values of x?
Now, similarly to that, what is the range of $(k-8)^2$ for all values of k?

What if you add 12 to this range now? So finally, for what values of k is $(k-8)^2+12>0$?

 

[TyErd]

im not sure but when you take a square onto the other side does the sign become opposite? like does < become >

 

[Mark44]

ohk, so (k-8)^2+12>0
(k-8)^2>-12

then what would do since you cannot square root a negative

You don't need to. The question you're trying to answer is "for what k are there two distinct, real solutions to the equation way back in this thread." There will be two distinct and real solutions of the equation x^2 + (8 - k)x - 3 = 0, if the discriminant (k - 8)^2 + 12 is positive.

So let's explore that:
(k - 8)^2 + 12 > 0 <==> (k - 8)^2 > -12

Since (k - 8)^2 >= 0 for any value of k, it will automatically be larger than -12 for any value of k.

So what's the answer to the question you're trying to answer: for what k are there two distinct, real solutions to the equation x^2 + (8 - k)x - 3 = 0?

 

[willem2]

Another way of doing this:

If the point (0,3) is inside the parabola y = x^2 - 8x then any non-vertical line through it must intersect it twice.

 

[TyErd]

the answer is k>8+2sqrt3 or k<8-2sqrt3

 

[Mentallic]

How did you get that?

Lets test it: For 2 distinct values cutting the line and the parabola, $(k-8)^2+12>0$. But you say that the solutions are $k<8-2\sqrt{3}$ and $k>8+2\sqrt{3}$. So for $8-2\sqrt{3}\leq k \leq 8+2\sqrt{3}$ it doesn't satisfy the inequality?
Well, let's try for an easy number, like 8.

$$(8-8)^2+12=12>0$$

this is an obvious contradiction to your solution, so it is obviously wrong.


You have ignored both mine and Mark44's attempt to help you understand the problem, without needing to do the usual manipulation of algebraic equations like you've become so accustomed to.
Open up your mind for a second, and forget about solving for k.

For now, all I ask you is this: when you put any number for x into $x^2$, can you ever get a negative number?