When does $x^2+(3k-7)x+(2k+6)=0$ have real roots?

When does $x^2+(3k-7)x+(2k+6)=0$ have real roots?

Question

Find for what values of $k$ the equation $x^2+(3k-7)x+(2k+6)=0$ will have real roots in $x$.

Suggestion

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The red curve here is $y=x^2+(3k-7)x+(2k+6)=0$.

What tells us how many real roots a quadratic equation has?

What is the discriminant of a quadratic equation?

How does its value determine how many roots of a quadratic are real?

Further suggestion…

Solution

A quadratic equation $ax^2 + bx + c = 0$ has real roots if and only if its discriminant $b^2-4ac \geq 0$.

Substituting in the quadratic’s coefficients, we find that for real roots here, we must have

$$\begin{align*} &&(3k-7)^2 -4(2k+6) &\geq 0 &&\quad\\ \iff\quad&& 9k^2 -42k + 49 -8k -24 &\geq 0 \\ \iff\quad&& 9k^2 -50k + 25 &\geq 0 \\ \iff\quad&& (9k-5)(k-5) &\geq 0. \end{align*}$$

The solutions of the quadratic equation $(9k-5)(k-5)=0$ are $k=\frac{5}{9}$ and $k=5$, so we can sketch the graph of $y=(9k-5)(k-5)$:

Therefore, the equation has real roots when $y\ge0$, which is when $k\geq 5$ or $k \leq \dfrac{5}{9}$.