Find the range of values for k such that ${kx^2 + 3x + 9k = 0}$ has real roots

Question

I am asked the question:

Find the range of values for ${k}$ such that ${kx^2 + 3x + 9k = 0}$ has real roots.

So from my understanding, there are distinct roots if ${b^2 - 4ac\ge 0}$

My first step is to substitute the values into ${b^2 - 4ac}$ which gives me:

${(3)^2 - 4(k)(9k)} \ge 0$

${9 - 36k^2} \ge 0$

divided by 9 becomes

${1 - 4k^2 \ge 0}$

=> ${1 \ge 4k^2}$

=> ${{1\over4} \ge k^2}$

=> ${{1\over 2} \ge k}$

=> ${k \le {1\over2}}$

Is my answer correct or have I taken a wrong turn?

Answer 1

Notice, you should use mod ($|\ \ \ |$) as follows $$k^2\le \frac{1}{4}$$ or $$|k|\le \frac{1}{2}$$ or $$-\frac 12\le k\le \frac 12$$

Answer 2

$k^2\leq \frac{1}{4}$ should give you $$-\frac{1}{2}\leq k\leq \frac{1}{2}$$

Answer 3

If $k<0$ we don't have $\sqrt{k^2}=k$. If $k<0$ we have $\sqrt{k^2}=-k$. For example: for $k=-2$ we have $$\sqrt{k^2}=\sqrt{(-2)^2}=\sqrt{4}=2=-k$$ Why? Because, by definition, the square root is always a non negative number.

Since you didn't cover modulus (absolute value) yet, you have to divide your solution in two steps: find the non negatives values of $k$ first and then find the nagatives values of $k$.

• For $k\geq 0$, we have: $$\vdots$$ $$\frac{1}{4}\geq k^2$$ $$\frac{1}{2}\geq k$$

• For $k< 0$, we have: $$\vdots$$ $$\frac{1}{4}\geq k^2$$ $$\frac{1}{2}\geq -k$$ $$-\frac{1}{2}\leq k$$

Conclusion: $\displaystyle -\frac{1}{2}\leq k\leq \frac{1}{2}$