Question

f(x)=x^2+(k+3)x+k
where k is a real constant.

Show that, for all values of k, the equation f(x)=0 has real roots.

Answer 1

k

Answer 2

f(x) = 0

x^2 + (k + 3)*x + k = 0

do you remember the solution to x in ax^2 + bx + c = 0, the quadratic formula

Answer 3

Like using the quadratic formula?

Answer 4

Or factoring?

Answer 5

f(x) is a quadratic, y = ax^2 + bx + c

a=1
b=k+3
c = k

Answer 6

The roots or zeros can be found by setting y = f(x) = 0 and solving for possible x values,

solving that general form of a quadrtic above for x, you get
\[x = \frac{ -b \pm \sqrt{b^2 - 4*a*c} }{ 2*a }\]

Answer 7

that is the quadratic formula,
notice the root term in the top...

Answer 8

that is what determines if you have real or complex roots,

Answer 9

For real roots, show that that quantity under the root is positive

Answer 10

If you have square root of a negative number, you get complex roots

Answer 11

complete the square $$f(x)=0\\x^2+(k+3)x+k=0\\x^2+(k+3)x+\frac{(k+3)^2}4=\frac{(k+3)^2}4-k\\\left(x+\frac{k+3}2\right)^2=\frac{(k+3)^2}4-k$$
so real roots \(x\) will exist where \(\left(x+\frac{k+3}2\right)^2\ge0\)

Answer 12

so... $$\frac{(k+3)^2}4-k\ge 0\\(k+3)^2-4k\ge 0\\k^2+6k+9-4k\ge 0\\k^2+2k+9\ge 0\\k^2+2k+1+8\ge 0\\(k+1)^2+8\ge 0$$ after completing the square again

Answer 13

now since \((k+1)^2\ge 0\) holds for all real \(k\) it follows that \((k+1)^2+8\ge 8\ge 0\) holds for all \(k\) as well and thus $$\frac{(k+3)^2}4-k\ge 0$$does too, meaning there is always a real root for any real \(k\)

Answer 14

I'm so sorry for wasting your guys' time. I had already completed the square but read it incorrectly and kept on getting negatives so I got frustrated and came here. Now that I looked back at it, I actually was doing it correctly... :D -> D:<

so sorry...

Answer 15

I'm so sorry for wasting your guys' time. I had already completed the square but read it incorrectly and kept on getting negatives so I got frustrated and came here. Now that I looked back at it, I actually was doing it correctly... :D -> D:<

so sorry...