# Find the other root of quadratic when $b^2-4ac = 0$?

According to many sources, the fundamental theorem of algebra states that every polynomial of degree $$n$$ has exactly n roots. But where's the other root when $$b^2-4ac = 0$$?

What's the other root of $$4x^2 - 32x + 64$$, for example? (the real root is 4).

• I just want to say that you're right: the given polynomial has exactly one root, not two. When counting roots with multiplicity, the polynomial has two roots (agreeing with its degree, as per the fundamental theorem of algebra), but when counting actual numbers that make the expression $0$, there is exactly one. – Theo Bendit Apr 24 at 5:54
• The most bare-bones statement of the Fundamental Thm of Algebra is that a non-constant polynomial with complex coeffs has at least one complex root; ie, we can "peel off" a factor: $p(z)=(z-r)q(z)$, where $q$ has lower degree than $p$. ... Now, we can continue, peeling-off factors until "$q(z)$" becomes constant; clearly, that'll take degree-of-$p$ steps altogether. Because we can encounter the same root/factor more than once in the process, the number of steps need not match the number of distinct roots. That's why it's important to say that the degree counts roots "with multiplicity". – Blue Apr 24 at 13:14

## 4 Answers

When $$b^{2} - 4ac = 0$$, the quadratic formula becomes $$-\frac{b}{2a}$$. You are wondering about the other root. This is where the concept of repeated roots/multiplicity would come in. The second root is equal to the first root, that's why you get only one value from the quadratic formula.

In your case, you have $$4x^{2} - 32x + 64 = 0$$. Notice that the left hand side can be factored into $$4(x - 4)^{2} = 0$$. In that case, by applying the zero property of multiplication, you get \begin{align*}x - 4 &= 0 &\qquad x - 4 &= 0 \\ x &= 4 & \qquad x &= 4.\end{align*} You can see that the root $$x = 4$$ is repeated twice.

Or, just by looking at the expression, you know that $$(x - 4)$$ is a factor which gives a root of $$x = 4$$. In your case, it is squared $$(x - 4)^{2}$$, that's why the number of roots increased, although the roots are the same.

The fundamental theorem of algebra requires roots to be counted with multiplicity. In conjunction with the factor theorem it implies that every univariate polynomial with complex coefficients can be broken into linear factors (of the form $$x-a$$, corresponding to root $$a$$); the multiple roots are the ones appearing more than once.

Here $$4x^2-32x+64=4(x-4)^2$$ and the root $$4$$ appears twice; it is a double root.

I believe the root is $$4$$ but occurs twice. This occurs when you can factor the expression as $$(x-4)^2=0$$. In this case, the discriminant of the quadratic is zero. So there will only be one solution.

The root $$x=4$$ is a repeated root. It can easily be proven that it is repeated if we solve the equation by completing the square:

$$4x^2-32x+64 = 0$$

$$(2x-8)^2 = 0$$

Take the square root on both sides:

$$2x-8 = ±\sqrt{0}$$

$$2x-8 = ±0$$

$$2x-8 = 0 \lor 2x-8 = -0$$

By moving the 8 to the other side, it's easy to see that the first root is the solution of the linear equation $$2x = 8+0$$ and the second root is the solution of the linear equation $$2x = 8-0$$

$$2x = 8+0 \Rightarrow 2x = 8 \Rightarrow x = 4$$

$$2x = 8-0 \Rightarrow 2x = 8 \Rightarrow x = 4$$

Both roots are 4 so the root is repeated. This is true for every quadratic polynomial where the discriminant is equal to zero.