Algebra 2 help

Question

Do you remember learning about the determinant? For a quadratic equation of the form: ax^2+bx+c

The determinant is equal to: b^2-4ac

It tells us how many solutions a quadratic equation has. The determinant can give us three results regarding the roots of an equation.

a) What are these three different results? What must the determinant be in order to get each result? Describe what a graphed quadratic equation would look like with each result.
b) Where does the determinant come from?
c) Use the ideas from parts A and B to prove that every quadratic polynomial has exactly two roots.

Answer 1

The question is relatively long to be answered here. Just take any textbook and read this topic. If you would have questions after that, you can ask.

a) By anyway, D=b²-4ac, determinant; three cases.

i) D>0. In this case, there are two distinct real solutions (or roots) of the equation (ax²+bx+c=0). In this case, graph the intersects x axis at two points x1 and x2, roots. The shape of the graph depends on D and coefficient a.

ii)D=0 two identical real solutions (basically one solution). x-axes is a tangent to the graph at the point  (x_12, 0).

ii)D<0 no real solutions. but two complex roots. In this case no intersection with x-axis.

 

b)Where b²-4ac came from?

Think of simplest quadratic equation, say x²=1. How would solve it? Just square root of 1 x=±√1. Other examples, like (x-1)²=4 or more 2(x+3)²=18 are solved in a similar way. Okay, if we can solve these equations. can we write equation ax2+bx+c=0 in that form? Yes we can.

ax²+bx+c=0

a[(x+b/2a)²-b²/4a2+c/a]=0

(x+b/2a)²=(b²-4ac)/4a²

x=[-b±√(b2-4ac)]/2a