# Homework Help: What value of k gives no roots,one root, 2 roots?

1. Nov 29, 2004

### aisha

roots! HELP

For what values of k does the equation x^2+k=kx-8 have two distinct roots, one real root, no real roots?

convert into standard form first well x^2+k-kx+8
I dont know if i can simplify this further? and if i cant then what does a=? b=? and c=? I dont understand how to do this plz help someone :uhh:

Last edited: Nov 30, 2004
2. Nov 30, 2004

### ShawnD

Group it differently and throw in some brackets
x^2 - kx + (k + 8)
a = 1
b = -k
c = (k + 8)

$$x = \frac{-b +- \sqrt{b^2 - 4ac}}{2a}$$

What determines the number of roots is the square root part

$$\sqrt{b^2 - 4ac}$$

If it's 0, there is 1 root. If it's an imaginary number, there are no real roots. If it's not 0 and it's real, there are 2 roots. Your condition for it equalling 0 (1 root) would be this:
b^2 = 4ac
(-k)^2 = 4(1)(k + 8)
k^2 = 4k + 32
k = -4, 8

Having 2 points to satisfy the condition that the square root term is 0 will create 3 ranges of k values with different results; k < -4, -4 < k < 8, and k > 8. Instead of trying to do some complicated math to figure out when the square root term will be imaginary, just take a test number from each of the 3 regions to determine what kind of numbers come from that region. Try the numbers k = -5, k = 0, and k = 10.

For k = -5:
square root {(-5)^2 - 4(1)(-5 + 8)}
=square root {25 - 4(3)}
=square root {25 - 12}
=square root {13}
=~ 3.6

For k = 0
square root {0^2 - 4(1)(0 + 8)}
=square root {0 - 32}
=square root {-32}
= imaginary number

For k = 10
square root {(10)^2 - 4(1)(10 + 8)}
=square root {100 - 4(18)}
=square root {28}
=~ 5.3

So now you know when the square root term is 0, you know when it's imaginary, and you know when it's not 0 and not imaginary. What can you conclude?

Last edited: Nov 30, 2004
3. Nov 30, 2004

### aisha

Does anyone else know how to do this problem? I dont think I understand what shawnD is saying.

4. Nov 30, 2004

### Nylex

What bit of what he said don't you get?

5. Nov 30, 2004

### aisha

Basically I dont understand what the values came out to be when k had no roots k had 2 real roots and k had one double root. I understand the concept and im also not sure why he grouped the numbers the way he did and how that affects the outcome, so will there be more values for k?

6. Nov 30, 2004

### Nylex

Do you understand the discriminant? For b^2 - 4ac > 0, you get real distinct roots, for b^2 - 4ac < 0, you get complex roots and for b^2 - 4ac = 0, you get a repeated root. I'm not sure why that is, but it's just like that.

As for grouping the equation, ie. x^2 - kx + (k + 8):

You know that a general quadratic has the form ax^2 + bx + c, where a, b and c are constants. You also know what the solutions to the equation ax^2 + bx + c = 0 are, in terms of those constants. Since you know that k is a constant, you should be able to see what a, b and c are by comparing ax^2 + bx + c with x^2 - kx + (k + 8).

7. Dec 1, 2004

### Diane_

Remember, aisha, a quadratic will always give you two roots. The two may be the same number, but there will still be two of them.

Note that k^2 = 4k + 32 is quadratic in k. There are, therefore, two k's that will solve it.

Try this:

k^2 = 4k + 32

k^2 - 4k - 32 = 0

(k - 8)(k + 4) = 0

k - 8 = 0 or k + 4 = 0

k = 8 or k = -4

There are two numbers because it turns out that two values will make the discriminant equal to 0, which is the condition for a single real root.

Side note, to be ignored if it doesn't make sense: I said there were always two roots, even if it's the same number. This is one such case: you have two roots, but you get one by adding zero and the other by subtracting zero. They end up being the same number. This may make more sense if you look at the problem geometrically. And, as I said, ignore this entirely if it confuses you.

8. Dec 1, 2004

### aisha

First of all in shawnD's solution what did the three values k<-4, -4<k<8, and k>8 mean? From what I gathered in ShawnD's solution I got that when K<-4 and<8 then there are 2 real zeros. When K>-4 but <8 there are no roots, and when K>-4 and >8 there are also 2 real roots. I dont understand what was the value of k when there was only one double root? I dont understand how to go about doing this question. Can someone plz thoroughly explain to me plz

9. Dec 2, 2004

### Astronuc

Staff Emeritus
Going back to what ShawnD's first post -

$$x = \frac{-b +- \sqrt{b^2 - 4ac}}{2a}$$

is the solution to the general quadratic equation:

$$a\,x^2\,+\,b\,x+\,c\,=0$$

ShawnD simply took your original equation and rearranged corresponding terms to get -

10. Dec 2, 2004

### aisha

I get that much I need more help. I dont understand what are the values of k when there are no zeros 2 zeros and 1 zero, I tried to understand what shawn did but since he picked random numbers -5,0, 10 i got confused. read the post b4 this I explain what I dont get, Please help

11. Dec 2, 2004

### shmoe

Take a look at the quadratic formula,

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Pay close attention to the $$\pm\sqrt{b^2-4ac}$$ in it. If the discriminant is positive, the root is a real number. You get one root for the + and one root for the -. If the discriminant is zero, this square root is zero and the "two" roots are given by

$$x = \frac{-b \pm 0}{2a}$$

In this case the plus or minus yields the same answer, so we get one root (in this case it's usually called a "double root"). If the discriminant is negative, this square root is a complex number, and we again get two roots, one for the + and one for the 1.

To summarize,
1) two real roots if $$b^2-4ac>0$$
2) one real root if $$b^2-4ac=0$$
3) no real roots (both are complex) if $$b^2-4ac<0$$

So the number of real roots is entirely determined by whether the discriminant is positive, zero or negative.

In this case, $$b^2-4ac=k^2-4k-32=(k-8)(k+4)=f(k)$$

This is a continuous function in the variable k, (labeled f(k) for convenience). So we just need to determine where f is positive, zero, or negative. We easily see it's zero when k=-4 or 8 easily enough, these will be the points where we have 1 root.

What is it's sign when k<-4? Since f is continuous, it's either positive or negative on any interval where it has no zeros. $$(-\infty,-4)$$ is such an interval. So, to find its sign on this interval it's enough to determine its sign at any point on the interval. Shawn took -5, but you could take any number you like that's less than 4. We see f(-5)=(-13)(-1)=13, which is positive, so f is positive when x<-4 and we therefore have 2 real roots when x<-4.

To see what the sign of f is if -4<x<8, you can again substitute any point on this interval you like, since f has no zeros here and is continuous, this is where Shawn's 0 came from. Likewise the 10 was to see what happens when x>8.

You might want to graph f(k) at this point, to get a better idea of what's going on.

12. Dec 2, 2004

### primarygun

k accompanies with a x term.

13. Dec 2, 2004

### Nylex

I can't believe I'm so stupid . Methinks I might have known that, it was just an off day or something . Thanks shmoe!

14. Dec 2, 2004

### Tom Mattson

Staff Emeritus
I'd like to thank everyone for their contributions, and at the same time remind everyone that we have a prohibition against posting complete solutions to homework problems. Students are expected to show how they started and where they got stuck, and then we help by giving them a little nudge.

Carry on,

15. Dec 2, 2004

### aisha

Ok THANKS SOOOO MUCH EVERYONE!!! I HOPE I GOT IT, lol. When k=8, and -4 then there is one zero, when k<8 but >-4 then there are no real roots. When k<-4 and >8 then there are two real zeros.

U guys tried so hard to explain but the algebra kept going over my head, I got these answers by sketching a graph with the zeros -4 and 8 and then just looked at the values as they became negative and positive? I hope this is correct? Thanks soo much Im going to need a little more time to absorb all ur explanations.