y = ax^2 + b+c?

Question

When using the quadratic formula to solve a quadratic equation (ax2 + bx + c = 0), the discriminant is b2 - 4ac. This discriminant can be positive, zero,
or negative.

Create three unique equations where the discriminant is positive, zero, or negative. For each case, explain what this value means to the graph of
y = ax2 + bx + c.

Answer 1

I wont give you examples but I will explain what happens in each case: positive discriminate means that there are two distinct roots; negative discriminate means there are no roots; zero discrinate means there is one double root.

Answer 2

well, lets start with positive:

b^2-4ac>0

b^2>4ac

if b=6, the b^2=36

if a=1, and c=3, 4ac=12

36>12, and 36-12=24>0

therefore ax^2+6bx+3=0 is an equation that gives you a positive discriminant.

if b^2-4ac=0

then b^2=4ac

if b=6 again, then b^2=36

if a =1, and c=9, then 4ac=36

36=36, and 36-36=0

therefore, ax^2+6bx+9=0 is an equation that gives you a discriminant equal to zero.

if you need b^2-4ac<0

then b^2<4ac

if a=1 again, b=2, and c=2

then 4<8, and 4-8= -4<0

therefore, ax^2+2x+2=0 is an equation with a negative discriminant.

ps. do you know what happens when you have a negative under the radical? like, the sqrrt(-4)? you introduce i! sqrrt(-4)=2i ! But don't let me confuse you, if you don't already know that, you will soon...

Answer 3

Generally, you start off by making a=1 when you look at quadratics. Hopefully you've already studied those in depth! Have you learned factoring and seeing that the factors are x-intercepts of the graph?

Try some numbers out.
How about a=1, b=6, c=9? y= x^2 +6x +9
How about a=1, b=5, c=6? y= x^2 +5x +6

and how about a=1, b=0, c=1? y= x^2 +0x +1 = x^2 + 1

You could graph them on your graphing calculator after you find the discriminants.

Answer 4

There are two distinct real solutions if the discriminant b2 – 4ac is positive (y=x^2+5x+3), one double real solution if the discriminant is 0 (y=4x^+4x+1), and no real solutions if the discriminant is negative (y=5x^+x+10)

Check my work before handing in your homework LOL

Answer 5

x^2-5x+4=0(1)
x^2-5x+25/4=0(2)
x^2-5x+24=0(3)
(1) here discriminant=25-16=9
The roots are positive
Also since it is perfect square they are rational
(2)Here it is equal to zero.
Therefore the roots are real and equal
(3)Here it is negative.so root are imaginary
because roots are={-b+-(b^2-4ac)^1/2}/2a

Answer 6

y = x^2 + 4x + 4
zero value, one answer
crosses at the x-axis once

y = x^2 - 1
2 real values, one negative and one positive
crosses at the x-axis twice

y = x^2 + 5x + 7
2 imaginary values, one negative and one positive
doesn't cross at the x-axis