check my answer please
- Thread starter haliebre
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We don't know: you haven't told us what the instructions were. Are you solving, and forgot to tell us what was on the other side of the "equals" sign? Are you find the zeroes? The vertex? Putting this in completed-square form?haliebre said:
Please reply with the full text of the exercise. Thank you.
Eliz.
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Okay then: what was the posted expression equal to?haliebre said:
Also, you posted "x2-4/3x +4/9". This could mean the following:
. . . . .\(\displaystyle \large{x^2\,-\,\frac{4}{3x}\,+\,\frac{4}{9}}\)
Is that what you meant? Or did you mean something like one of the following?
. . . . .\(\displaystyle \large{\frac{x^2\,-\,4}{3x}\,+\,\frac{4}{9}}\)
. . . . .\(\displaystyle \large{x^2\,-\,\frac{4}{3}x\,+\,\frac{4}{9}}\)
When you reply with what the expression was equal to, please include all the steps you did in the solving process. Thank you.
Eliz.
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So the original exercise was as follows:haliebre said:
. . . . .Solve the following:
. . . . .\(\displaystyle \large{x^2\,-\,\frac{4}{3x}\,+\,\frac{4}{9}\, =\,0}\)
That still leaves us wondering how you obtained solutions of x = 4/3 and x = 1/3, since:
. . . . .x<sup>2</sup> - 4/(3x) + 4/9 = 0
. . . . .[4/3]<sup>2</sup> - 4/(3[4/3]) + 4/9 ?=? 0
. . . . .16/9 - 1 + 4/9 ?=? 0
. . . . .20/9 - 1 ?=? 0
. . . . .20/9 - 9/9 = 11/9 ?=? 0
So "x = 4/3" doesn't work. And:
. . . . .x<sup>2</sup> - 4/(3x) + 4/9 = 0
. . . . .[1/3]<sup>2</sup> - 4/(3{1/3]) + 4/9 ?=? 0
. . . . .1/9 - 4 + 4/9 ?=? 0
. . . . .5/9 - 4 ?=? 0
. . . . .5/9 - 36/9 = -31/9 ?=? 0
So "x = 1/3" doesn't work.
Please reply showing your steps. For instance, first you multiplied through to clear the denominators. What did you multiply by? What did you get? What did you do next?
Thank you.
Eliz.