Solving Quadratic Equations By Graphing....

[Guest]

Allright then. This is giving me a real hard time. I listen take notes and it's just not helping.

Slove each equation by graphing. If exact root cannot be found, state the consecutive integers between which roots lie.

4x^2-12=0.... Allright, I tried to factor it, thats what my teacher said. And I got x=-4 and x=3.. Then he said to use x=b/2(a).. And I just go stupid there :shock: . I don't get it. If you can help, I would really really like it. Thank you.
Brittany

 

[pka]

Look at the instructions!
Did you graph the function?
What are the x-intercepts?
Those are your roots!

 

[stapel]

[Solve] each equation by graphing.

Have you done the graph? If so, what did it show you?

If exact root cannot be found, state the consecutive integers between which [the] roots lie.

4x^2-12=0.... Allright, I tried to factor it, thats what my teacher said.

So you have learned only how to solve by factoring...? No other methods? How do you find non-integral roots, then? Please clarify. Thank you.

And I got x=-4 and x=3.

Factoring should have produced factors, not x-values. Where did these x-values come from? Please reply with all of your steps. Thank you.

Then he said to use x=b/2(a).

So you're not actually solving...? You're actually finding the vertex of the associated parabola...?

Please reply with the full and exact text of the exercise, the complete instructions, and a clear listing of all of your steps so far. Thank you.

Eliz.

 

[Mrspi]

Allright then. This is giving me a real hard time. I listen take notes and it's just not helping.

Slove each equation by graphing. If exact root cannot be found, state the consecutive integers between which roots lie.

4x^2-12=0.... Allright, I tried to factor it, thats what my teacher said. And I got x=-4 and x=3.. Then he said to use x=b/2(a).. And I just go stupid there :shock: . I don't get it. If you can help, I would really really like it. Thank you.
Brittany

Hmmmm......
You have

4x<SUP>2</SUP> - 12 = 0

Divide both sides of the equation by 4:

x<SUP>2</SUP> - 3 = 0

If you are looking for the solutions for this, then you can factor the left side:
(x + sqrt(3))(x - sqrt(3)) = 0

x + sqrt(3) = 0, or x - sqrt(3) = 0
x = - sqrt(3) or x = sqrt(3)

If x = - sqrt(3), then x lies between -2 and -1.
If x = sqrt(3), then x lies between 1 and 2.

 

[Guest]

Well thanks for all your help everybody. I had a test today. But I don't think I did good. But I can re-test. And I don't think we are doing this stuff anymore. Yay. Well I'll probly be on again for help. Thanks.
Brittany

 

[Gene]

Not to beat a dead horse but the vertex is at
x = -b/(2a) and in this equation b=0.
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Gene