Find the range of values k can take for $kx^2 + 2x + 1 - k = 0$ to have two real distinct roots

Question

I'm stuck on this question, and I have the answer but I don't know how to get to it. This is what I've done so far.

For the quadratic equation $ax^2+bx+c=0$ to have two real roots, the discriminant must be greater than $0$: $$ b^2 - 4ac > 0. $$ After substituting $a$, $b$ and $c$ in, I got $$ 2^2 - 4k(1-k) > 0. $$ And then after expanding and simplifying, I got $$ k^2 - k + 1 > 0. $$ I can't solve this, so how would I work out the range of values that $k$ can be?

This was a textbook question, and the answer was $$ \{k \text{ is a member of all real numbers such that $k$ does not equal 0}\}. $$ I don't understand why this is the answer, can someone explain how we get there please?

Answer 1

You have done most of the work already. What's left is to note that, as a function of $k$, $k^2-k+1$ is a parabola with minimum value $3/4$ for $k=1/2$. In other words, $k^2-k+1 > 0$ for all $k$. However, we need to discard $k=0$, since, for that value, your original equation reads $2x+1=0$, which only has one root.

Answer 2

the discriminant of k is 1-4=-3<0 and the coefficient of k^2 is 1 so the expression is positive for all values of k. The above and the fact that k =/= 0 since the original equation must be quadratic gives the textbook answer.

Answer 3

Let me show you a beautiful way to solve your issue:

$$k^2-k+1 > 0$$

This is equivalent to:

$$\frac{k^3+1}{k+1} > 0$$

The sign of a division of two equations is equal to the product of those equations (taking into account that the denominator is not zero, so it comes down to:

$$(k^3 + 1) \cdot (k + 1) > 0$$

This is true if both are either negative or positive. Now we are very lucky, because taking the third power (or root) does not change the sign:

$$k + 1 > 0 \iff k > -1 \iff k^3 > -1 \iff k^3 + 1 > 0$$

In other words: $k+1$ and $k^3+1$ are either both positive or both negative, so their product or quotiënt is always positive.

Now what about the special case that $k+1$ (and $k^3+1$) are zero ($k=-1$)?
Well, in that case you end up with $k^2-k+1$ being $1$, which is larger than zero, so that should also yield an equation with two different roots, and indeed: $-x^2+2x+2=0$ is a second-degree equation with two different roots.