# Homework Help: Find k probably too easy

1. Apr 1, 2009

### King

Hi all,
I have just an easy maths question which my brain is incapable of calculating for some reason...

How would you find the value of k in: x^2 + 10x + k = 0?

I have come up with (x+5)^2 +25=0, so is k=25? Is this correct?

My maths knowledge has declined greatly in the duration of about 6 months...

Thanks for any help :)

2. Apr 1, 2009

### Staff: Mentor

I suppose you are not just looking for k, but for k such that the equation has 1 or 2 roots?

3. Apr 1, 2009

### thrill3rnit3

Based from your given data, K can be....anything??

Maybe you missed some other restriction, say, that the polynomial is a perfect square or something.

4. Apr 1, 2009

### King

Oops sorry, I forgot to mention that the equation has equal roots.

5. Apr 1, 2009

### JG89

If the quadratic has equal roots then the vertex will have to touch the x-axis. You can write your quadratic in this form: (x - c)^2 + d = 0, where c and d are some real numbers (which satisfy the equation, of course). Try completing the square and see if you can figure out the value of k to make the vertex touch the x-axis.

6. Apr 1, 2009

### JG89

Oops, you already have the right answer. You shouldn't get (x+5)^2 +25=0 though. You should have (x+5)^2 - (25 - k).

7. Apr 1, 2009

### deancodemo

If the quadratic equation ax^2 + bx + c = 0 has equal roots, then the descriminant = 0. That is, b^2 - 4ac = 0.

Applying this to your question, we have:
100 - 4 * 1 * k = 0
100 - 4k = 0
4k = 100
k = 25.

8. Apr 1, 2009

### meiso

Another way you could solve this, just for kicks.

x^2 + 10x + k = 0

Corresponding to the coefficients of the standard form of a quadratic equation,

a = 1
b = 10
c = k

r1 = root one
r2 = root two

since the roots are equal, r = r1 = r2

The sum of the roots:

r + r = -b/a
2r = -10/1
r = -5

The product of the roots:

r * r = c/a
r^2 = k/1
(-5)^2 = k
k = 25

9. Apr 2, 2009

### King

Thanks for the responses, problem solved