For what interval of $k$ does the equation have one postitive and one negative root?

Question

I was given an equation that looked like this

$(k-x)(1-x) + 4$

and was told to find the interval of k when the equation would have one positive root and one negative root.

So far I have found the solutions in terms of k

$x_1 = \frac{1}{2}(k+1-\sqrt{k^2-2k-15})$

$x_2 = \frac{1}{2}(k+1+\sqrt{k^2-2k-15})$

I am a bit stuck of where to go from here, do we have to figure it out using $\sqrt{b^2-4ac}$?

Answer 1

Let $f(x)=(x-1)(x-k)+4=x^2-(k+1)x+(k+4)$. So $y=f(x)$ represents a parabola facing up (concave up). For the roots to be of opposite signs and real, we need $$f(0)<0 \quad \text{ and} \quad \text{discriminant } \geq 0.$$ Thus $k+4 < 0$ and $(k+1)^2-4(k+4) \geq 0$. But the latter inequality is true if the first inequality holds. So $k <-4$.

Answer 2

Expand your equation to get $$x^2-(k+1)x+k+4=0$$

For this equation to have one positive and one negative root, the product of its roots is negative, thus $$k+4<0$$

$$\therefore k < -4$$

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You don't have to check $D=b^2-4ac$, since the product of roots of the quadratic equation $ax^2+bx+c=0$ is $c/a$, but since this value is negative, $ac$ is also negative, thus resulting in $D \geq 0$

Answer 3

I made some silly errors in my original answer. The corrections now seem to show that k < -4 is the condition.

Following dxiv's advice, the roots are $x_1 = \frac{1}{2}(-\sqrt(k^2-2k-15)+k+1) $ and $x_2 = \frac{1}{2}(\sqrt(k^2-2k-15)+k+1) $.

For there to be two real roots, we must have $k^2-2k-15 > 0$ or $k^2-2k+1 > 16$ or $(k-1)^2 > 16 $ or $k-1 > 4$ or $k-1 < -4$ or $k > 5$ or $k < -3$.

(Error - had -5 here)

To make the roots of different signs, consider the cases separately.

If $k > 5$ then $\sqrt(k^2-2k-15)+k+1 > 0$ so we want $-\sqrt(k^2-2k-15)+k+1 \lt 0$ or $\sqrt(k^2-2k-15) \gt k+1 $ or $k^2-2k-15 \gt k^2+2k+1 $ or $0 > 4k+16$ which never happens.

(I messed up the inequality below also)

If $k < -3$ then $\sqrt(k^2-2k-15)+k+1 > 0$ so we want $\sqrt(k^2-2k-15) \gt -k-1 $ or $k^2-2k-15 \gt k^2+2k+1 $ or $0 > 4k+16$ or $k <-4$,