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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?

dvdgrgrtt
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m4themat1cs
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  • Before any of that you need a quadratic equation, which yours is not. – Paul Nov 01 '23 at 09:15
  • Please edit the title - we understand that the equation should read $kx^2 +2x+1-k=0$, right? Actually, all your maths need to be properly formatted. – dvdgrgrtt Nov 01 '23 at 09:30

3 Answers3

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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.

Janek Szaniewski
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Nov 01 '23 at 09:17
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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.

dvdgrgrtt
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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.

Dominique
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