math help!?

Question

determine the value(s) of k that give the type of solution indicated.

x^2 + kx + 16 = 0 ; 1 real root

im stuck on this question and i need your help please! thank you!

Answer 1

When you have one real root, the quadratic equation touches the x-axis at only one point.

If given the general equation:

ax^2 + bx + c,

Then for this to happen,

b^2 - 4ac = 0

So, in your question,

x^2 + kx + 16 = 0,

we have,

a =1
b = k
c = 16

Substituting,

b^2 - 4ac = 0

(k)^2 - 4(1)(16) = 0

k^2 - 64 = 0

k = sqrt(64)

k = 8 or k = -8

Answer 2

You can use the quadratic formula, but don't need to. How can a quadratic have a single real root? Only if the root is of order 2. (imaginary roots would be in pairs, if there is 1 real root there must be 2, and so it must have multiplicity 2).

So you need something of the form
(x - r)^2 = x^2 + kx + 16
x^2 - 2rx + r^2 = x^2 + kx + 16
-2rx = kx and r^2 = 16
r^2 = 16 => r = 4 or -4, -2r = k, so k = 8 or -8.

As often happens, there is more than one way to do a problem.

Answer 3

In the quadratic formula you need the term "b^2-4ac" to be zero.

Therefore you need

k^2 - 4*16 = 0

or

k^2 = 4*16

or

k = +8 or -8 (two possible solutions)