Quadratic Functions...?

Question

Prove that the roots of the equation:

x^2 + (k + 2)x + 2k = 0

are real for all values of k.

I need help on this, the last set of answers where wrong,
all i know that-
Real roots: b^2 > 4ac
Anyone online that can help?

Answer 1

So just write down what you already wrote (only it is slightly wrong: it needs a >= instead of just a >):
The equation has real roots whenever b² >= 4ac

Now plug in and simplify:
The equation has real roots whenever
(k+2)² >= 4(1)(2k)
k²+4k+4 >= 8k
k²-4k+4 >= 0
(k-2)² >= 0

Since the term on the left is a square, it is always greater than or equal to 0. Thus, you will always have real roots.

Answer 2

For Real roots: b^2 > 4ac
=> (k + 2)^2 > 8k
=> k^2 + 4 + 4k > 8k
=> k^2 + 4 - 4k > 0
=> (k - 2)^2 > 0

Which is true for all K as (k - 2)^2 is a perfect square always greater than 0.

Hence the roots are real for all values of k.

Answer 3

x^2 + (k + 2)x + 2k = 0

In the above equation a+1 b = k+2 and c = 2k
So (k+2)^2 >8k
k^2+4k+4>8k
k^2 -4k+4 >0
(k-2)^2>0
The above is true for all k because the squared term is always positive and thus > 0.

Answer 4

The roots are (–2, –k). –2 is clearly real under any circumstance leaving only –k to consider. If k is real, then –k is clearly real, however, there is no constraint in the problem forcing k to be real.

Therefore, the roots of the equation are NOT real for all values of k.

Answer 5

If b^2 > 4ac then the values of x are real. The way to find the values of x is in accordance with the equation, namely:
x= -b+ or- sqrt(b^2 -4ac)/2a

x= -(k+2) plus or minus sqrt(k+2)^2 -4*2k)/2
x = -(k+2) plus or minus sqrt(k^2+4k+4-8k)/2
x= -(k+2) plus or minus sqrt(k^2-4k+4)/2
x= -(k+2) plus or minus sqrt{(k-2)^2}/2
x= -(k+2) plus or minus (k-2)/2
x = (-k-2 + k-2)/2 or (-k-2 -k+4)/2
x = -4/2 or -2(k+1)/2
x = -2 or -(k+1). For all values of k, x is a real value! to go on:
x^2 + (k + 2)x + 2k = {x+2}{x+(k+1)} =0
x= -2
x= -(k+1)
That is it!

Answer 6

well b = (k + 2), and b^2 = k^2 + 4k + 4
4ac = 8k

Hence you must prove that k^2 + 4k + 4 >/= 8k
ie k^2 -4k + 4 >/= 0
but the lhs = (k - 2)^2, which is always greater than or equal to 0

hence b^2 >/= 4ac