Finding $k$ such that the discriminant $k^2 - 12k + 52$ yields a quadratic with only one solution

Question

The given discriminant (not the quadractic) is $k^2 - 12k +52$

The exact question asks "find value of $k$ when there is only one solution" - that would mean the equation would equal $0$:

$$k^2 - 12k + 52 = 0$$

When I plot it as a graph, it doesn't hit the $x$ axis, so it would have $2$ imaginary solutions (and not one), which contradicts the question itself. I feel like I'm missing something here.

Answer 1

You've got things a little backwards.

So, what you have seen is that $k^2 - 12k + 52>0$ for all real $k$. This does mean that the quadratic $k^2 - 12k + 52$ has only two non-real solutions ... but that's not your concern.

Remember, this is still (what I believe you mean to call) the discriminant. Recall: if a quadratic has discriminant $D$, then

• $D>0$ means two, real, distinct solutions to the original quadratic
• $D=0$ means one real solution to the original quadratic (a double root)
• $D<0$ means two, non-real, distinct solutions to the original quadratic

This should make sense: the quadratic formula is essentially

$$x=\frac{-b \pm \sqrt D}{2a}$$

so:

• $D>0$ means you get a solution for $+\sqrt D$ and another for $-\sqrt D$
• $D=0$ means you just have $+0$ and $-0$, the same thing
• $D<0$ means that you have the square root of a negative

---

However, ultimately, our conclusion is the same, even if for the wrong reasons in your case.

The discriminant $k^2 - 12k + 52$ is always strictly positive. Consequently, the original quadratic, no matter what $k$ is (as a real number), we'll have two real, distinct solutions.

---

Emphasis on $k$ is a real number:

But what if $k$ is not a real number? Sure, it'd make the most sense if this is the kind of question you're being asked. (It's a reasonable question for an Algebra I class just learning of discriminants.)

But using the quadratic formula and imaginary numbers, you can happily find a $k$ such that $k^2 - 12k + 52 = 0$:

$$k = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(52)}}{2(1)} = \frac{12 \pm \sqrt{144 - 208}}{2} = \frac{12 \pm \sqrt{-64}}{2} = 6 \pm 4i$$

So if $k$ is allowed to be a complex number, then you can get a discriminant of zero still, and have only one real solution.

A basic quadratic with such a discriminant would be

$$f(z) = z^2 - \frac{k^2 - 12k + 52}{4}$$

albeit rather trivially. Since we choose $k$ such that $k^2 - 12k + 52 = 0$, then trivially the solution here is $0$ (as a double root).

Answer 2

As we all know, a quadratic:
$ax²+bx+c=0$
has two solutions,
$x=[-b+√(b²-4ac)]/2a$ & $[-b-√(b²-4ac)]/2a$.
These are only equal,i.e., the quadratic only has one solution when the discriminant is zero.
In this particular case, k being complex is completely fine. Solving for $k²-12k+52=0$, we get $k=6+4i$ & $6-4i$.