Solving Quadratic Equations by Graphing

ChickySeeChickyDo

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I am having so much difficulty understand this! I have had my teacher explain this and the tutor, and I still don't get it.

1) x²-9=0

How do I solve this? Please tell me step by step.
 
1) x²-9=0

How do I solve this?

Just replace the 0 with y, so the equation becomes

y = x^2 - 9

This graphs as a parabola with vertex at (0,-9) and x-intercepts at (-3,0) and (3,0). The x values at the intercepts are the solutions to the original equation. This is the graphical method of solving.

The algebraic method is to simply factor and solve:

x^2 - 9 = 0
(x + 3)(x - 3) = 0
x = 3 OR x = -3

Hope that helps.
 
\(\displaystyle x^2-9 \ = \ 0\)

\(\displaystyle x^2 \ = \ 9\)

\(\displaystyle \sqrt{x^2} \ = \ \sqrt9\)

\(\displaystyle |x| \ = \ 3\)

\(\displaystyle x \ = \ \pm3\)

\(\displaystyle See \ graph\)

[attachment=0:3ryrv9dj]zfg.jpg[/attachment:3ryrv9dj]
 

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BigGlenntheHeavy said:
\(\displaystyle x^2-9 \ = \ 0\)

\(\displaystyle x^2 \ = \ 9\)

\(\displaystyle \sqrt{x^2} \ = \ \sqrt9\)

\(\displaystyle |x| \ = \ 3\)

\(\displaystyle x \ = \ \pm3\)

\(\displaystyle See \ graph\)

[attachment=0:1yuyuxjb]zfg.jpg[/attachment:1yuyuxjb]

Thanks^^ The only thing is I just don't understand how to graph the points. And my answer comes in zeros.
 
\(\displaystyle f(x) \ = \ x^2-9\)

\(\displaystyle f(3) \ = \ 0\)

\(\displaystyle f(-3) \ = \ 0\)

\(\displaystyle Hence, \ solutions \ are \ (3,0),(-3,0) \ to \ graph \ of \ f(x) \ = \ x^2-9\)

\(\displaystyle Or \ manually \ graph \ f(x) \ and \ see \ where \ the \ graph \ intercepts \ the \ x-axis.\)

\(\displaystyle At \ that \ point \ y \ = \ 0, \ so \ we \ found \ a \ solution \ of \ f(x)\)
 
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