G
Guest
Guest
i have 2 equations that i need to reduce into lowest possivle form
29x+30^2-10=40
and
5x^2+20x=-5
29x+30^2-10=40
and
5x^2+20x=-5
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- Feb 4, 2004
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Fractions get "reduced to lowest terms". Equations get "solved". So I'm afraid your post, as it stands, doesn't make much sense.
Please reply with the full text of and instructions for this exercise (or are these two exercises?), showing what you have tried thus far.
Thank you.
Eliz.
Please reply with the full text of and instructions for this exercise (or are these two exercises?), showing what you have tried thus far.
Thank you.
Eliz.
G
Guest
Guest
yes they are 2 different problems...equations whatever you want to call them i just need to solve them!! i have not tried anything but
getting the 40 (1st equation) and the -5 f(rom the othe equation) to the other side!!!
getting the 40 (1st equation) and the -5 f(rom the othe equation) to the other side!!!
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Just move the numbers (by appropriate adding and subtracting), just like you did back when you were solving linear equations. The process is the same. Then simplify.
Oh, and I noticed that your subject line mentions what might be the instructions. Are you supposed to be solving the quadratic equations by using the Quadratic Formula? If so, then just plug the numbers into the formula you've memorized:
. . . . .For ax<sup>2</sup> + bx + c = 0,
. . . . . . . .\(\displaystyle \large{x\,=\,\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}}\)
Then simplify, if possible. (It isn't always possible, by the way.)
Eliz.
Oh, and I noticed that your subject line mentions what might be the instructions. Are you supposed to be solving the quadratic equations by using the Quadratic Formula? If so, then just plug the numbers into the formula you've memorized:
. . . . .For ax<sup>2</sup> + bx + c = 0,
. . . . . . . .\(\displaystyle \large{x\,=\,\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}}\)
Then simplify, if possible. (It isn't always possible, by the way.)
Eliz.
G
Guest
Guest
using one of my equations can you plug it in to the quadratic formula so i can see how that works?
i have used it before but it was a long time ago and i was 50% sure that is how you solve the equations i needed solved
i have used it before but it was a long time ago and i was 50% sure that is how you solve the equations i needed solved
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You just read the numbers off, and plug them in. For instance:
. . . . .\(\displaystyle \large{-5x^2\,+\,2x\,+\,7\,=\,0}\)
Clearly a = -5, b = 2, and c = 7, so:
. . . . .\(\displaystyle \large{x\,=\,\frac{-(2)\,\pm\,\sqrt{(2)^2\,-\,4(-5)(7)}}{2(-5)}\,= \,\frac{-2\,\pm\,\sqrt{4\,+\,140}}{-10}}\)
...and so forth.
Eliz.
. . . . .\(\displaystyle \large{-5x^2\,+\,2x\,+\,7\,=\,0}\)
Clearly a = -5, b = 2, and c = 7, so:
. . . . .\(\displaystyle \large{x\,=\,\frac{-(2)\,\pm\,\sqrt{(2)^2\,-\,4(-5)(7)}}{2(-5)}\,= \,\frac{-2\,\pm\,\sqrt{4\,+\,140}}{-10}}\)
...and so forth.
Eliz.
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Your answers are "(-29 ± sqrt[6841]) / 60" and "-2 ± sqrt[3]". I hereby apologize for previously having provided step-by-step instructions on how to arrive at the solutions. I would appreciate receiving no further obscenity-laden private messages.
Thank you for your consideration.
Eliz.
Thank you for your consideration.
Eliz.