**Write-Up #3 (Quadratic Equations)**

*ax ^{2}+ bx + c = 0*

By Jaepil Han

Inverstigation 1. Graph in the xb plane.Consider again the equation

x^{2}+ bx + 1 = 0Now graph this relation in the xb plane.

xis a equation in a certain situation, when the values of both^{2}+ bx + 1 = 0andaare 1. You might be interested in the differet roles of the parameters,canda. However, here's a thing. Even the value ofcis not 1, we can simply make the equation as a monic equation since the equation is divisible by the leading coefficienta. For example, when a = 2, the equation isa2x. But, this can be rewrote as^{2}+ bx + 1 = 0x. Eventually, this can be expressed as^{2}+ (b/2)x + 1/2 = 0x. For this reason, that is enough to investigate when the parameter^{2}+ b'x + 1/2 = 0is 1. Therefore, you may assume that all the investigations followed below are always based on the situations whena=1.a

First, let's see the graphs when the values of c are positive.

Here's the graphs of the equations for c = 1, 3, 5, 7, 9

By placing the horizontal lines, such as b=4, we can simply figure out there are two intersections between the hyperbola and the horizontal line. For the various b values,in addition, there are two points when there is a single root, and there is a region when there is no root. Thus, we can say the equation has at most two roots.

Second, let's see the equation when c=0

x^{2}+ bx = 0

bx = - x^{2}

b = - xHere's the graph of the equation.

Therefore, the equation is linear when the parameter

=0.c

Third, let's consider the values of c are negative.

Here's the graphs of the equations for c = -1, -3, -5, -7, -9

Similarly, when we place the horizontal lines, such as b = 4, we can figure out that there are always two intersections between the hyperbola and the horizontal line. One of the roots is positive, and the other is negative. We can also say that when we place the horizontal line far from the x-axis, one of the roots is close to the asymptote x=0, and the other root is close to the other asymptote x=-b.

Here's a video when the value of b varies from -5 to 5, the horizontal line, with the graph of the equation

, x^{2}+ bx + 1 = 0.

If the horizontal line is b=2 or -2, the horizontal line and the graph of the

xmeet at exactly one point.^{2}+ bx + 1 = 0

Algebraically,

x^{2}+ 2x + 1 = 0

(x + 1)^{2}= 0

∴ x=-1

x^{2 }- 2x + 1 = 0

(x -1)^{2 }= 0

∴ x=1Therefore, there are two points, which have a single root of the equation

x.^{2}+ bx + 1 = 0

If the horizontal line is greater than 2 or less than -2, the horizontal line and the graph of the

xmeet at two intersection points. In other words, they have two real roots. Probably, this may happen when c has positive values, like c = 1, 3, 5, 7, 9.^{2}+ bx + 1 = 0

Here's another video when the value of b varies from -5 to 5 as same as the previous one, but with the graph of the equations,

xand^{2}+ bx + 1 = 0x.^{2}+ bx - 1 = 0

The horizontal line and the graph of

xalways meet at two points, one has a negative x value and the other one has a positive x value. Also, this happens when c's are negative, like -1, -3, -5, -7, and -9. This means that^{2}+ bx - 1 = 0xalways have two different real roots.^{2}+ bx - 1 = 0

Now, let's think about the features of the roots. Interestingly, all the hyperbolas have two asymptotes x=0 and x=-b. So, here's a thing. We might have a conjecture that when the horizontal line is getting higher or lower, then the two roots of the equation

xare close to the asymptotes, x-0 and x=-b.^{2}+ bx + c = 0Algebraically, explicitly solve the equation for x (actually, it is the same as the quadratic formula)

Therefore, if the horizontal line b is extremly large or small, then the roots of the equation are very close to -b and 0. This may explain why one root approaches -b and the other one approaches to 0.