Difference between "two equal roots" and "one root"?

9.2k Views Asked by At

When solving a quadratic equation for which the discriminant is zero (ie. $b^2-4ac = 0$) we say there are "two real and equal roots". Why do we emphasise that there are two different but equal roots, as opposed to saying there is only one root? Is there any geometrical intuition for why it would be incorrect to think of these roots as one, or is it simply a convention?

2

There are 2 best solutions below

0
On BEST ANSWER

The fundamental theorem of algebra states that every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots.

By counting the multiplicity, we have this nicely stated theorem, which is useful for instance if you want to study the algebraic multiplicity of eigenspace.

0
On

Of course, multiplicity of roots of an equation is important and has a lot of related facts. For example when you factor a polynomial. But I am going to try to explain the difference geometrically for equations of second degree.

If you think of the equation $ax^2+bx+c=0$ as a function $y=ax^2+bx+c$ then you have a parabola ($a\ne 0$). The solutions of the equation are the points where the parabola intersects the $OX$ axis. What is the difference between two equal roots and two different roots? If you have two different roots then the parabola intersects the $OX$ axis at two different points. (In this case, it is $ax^2+bx+c=a(x-r)(x-s)$ where $r,s$ are the roots.) If there is only one root the parabola intersects the $OX$ axis at only one point (the vertex). (In this case, it is $ax^2+bx+c=a(x-r)^2$ where $r$ is the root. Note the $2$ in the exponent. That is the reason to say it has multiplicity two.) I think you can deduce what happens if the equation has no real solution.