Need math help again......please!!!!!!11?

Question

I need to find all real values of K that will make x^2+2Kx+20 a perfect square?

Answer 1

Note

In order for x^2 + 2kx + 20 to be a perfect square, it must be factorable as (x + ?)^2.

Remember that 20 is supposed to be "half squared" of the middle term. i.e. if b is the middle term,

20 = (b/2)^2
Now, we solve for b, to get
20 = b^2 / 4
80 = b^2, and b = +/- sqrt(80) = +/- 2 sqrt(20)

Now, the middle term is 2k, so 2k must equal 2 sqrt(20)
2k = +/- 2 sqrt(20)
k = +/- sqrt(20)

Therefore, all values of k that make that a perfect square are:
k = sqrt(20)
k = -sqrt(20)

(where sqrt = "square root")

Let's check.

Claim: x^2 + 2*sqrt(20)x + 20 and x^2 - 2*sqrt(20)x + 20 are perfect squares.
Proof: (x + sqrt(20))^2 = x^2 + sqrt(20)x + sqrt(20)x + 20
= x^2 + 2*sqrt(20)x + 20

Similarly, with k=-sqrt(20),
x^2 - 2*sqrt(20)x + 20 factors into (x - sqrt(20))^2

Answer 2

Probably the teacher expects you to find the vaule of 2K

x^2+2Kx+20

Perfect square=(x+a)^2 and 20 is A^2 in this case
Expands to
x^2+2ax+a^2

a=Square root of 20= 2(square root of 5)
K=(square root of 20)x
2K=4(square root of 5)x

So K= (square root of 20)

Answer 3

I am not sure if you are looking for 1) or 2)

1)
If 20 = k^2
then x^2+2kx+k^2 = (x+k)^2
So k = srqt(20) = 4.472.. or k = -4.472..

2)
The perfect squares: 0,1,4,9 etc. are obtained by squaring 0,1,2,3 etc.....or squaring -1,-2,-3, etc.
So we must have x+k=0,1,2,3,....or x+k= -1,-2,-3...
So, in terms of x, we must have k = -x, 1-x, 2-x, 3-x....
or k = -(1+x), -(2+x), -(3+x)......

Answer 4

Ultimately you want this to be in the form:
(x + a)²

So expand this out. The result will be:
x² + 2ax + a²

Now start matching this with your equation:
a² = 20
K = a

Taking the sqrt of the top equation:
a = sqrt(20)

Then notice how K = a = sqrt(20).

So K = sqrt(20).

The final squared form will be (x + sqrt(20))²