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.
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.
Just replace the 0 with y, so the equation becomes1) x²-9=0
How do I solve this?
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.
[tex]x^2-9 \ = \ 0[/tex]
[tex]x^2 \ = \ 9[/tex]
[tex]\sqrt{x^2} \ = \ \sqrt9[/tex]
[tex]|x| \ = \ 3[/tex]
[tex]x \ = \ \pm3[/tex]
[tex]See \ graph[/tex]
[attachment=0:3ryrv9dj]zfg.jpg[/attachment:3ryrv9dj]
I am not, therefore I do not think. Contrapositive of Descartes' quip.
Thanks^^ The only thing is I just don't understand how to graph the points. And my answer comes in zeros.Originally Posted by BigGlenntheHeavy
[tex]f(x) \ = \ x^2-9[/tex]
[tex]f(3) \ = \ 0[/tex]
[tex]f(-3) \ = \ 0[/tex]
[tex]Hence, \ solutions \ are \ (3,0),(-3,0) \ to \ graph \ of \ f(x) \ = \ x^2-9[/tex]
[tex]Or \ manually \ graph \ f(x) \ and \ see \ where \ the \ graph \ intercepts \ the \ x-axis.[/tex]
[tex]At \ that \ point \ y \ = \ 0, \ so \ we \ found \ a \ solution \ of \ f(x)[/tex]
I am not, therefore I do not think. Contrapositive of Descartes' quip.
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