- #1
danago
Gold Member
- 1,118
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Hey. Today i had a test on quadratics and discriminants. I think i did fairly well, but i am a bit confused about one of the questions i had in it.
We were given the following quadratic equation:
12x^2 + 8mx + (4m-3) = 0
What we had to do was prove that for any integer value of m, the equation would have rational solutions.
What i did was first take the discriminant of the equation:
\Delta = (8m)^2 - 4(12)(4m-3)
simplified it:
\Delta = 64m^2 - 192m + 144
From that, i then created a table of values. I made a table of values for m (-5 to 5), the discriminant, then the square root of the discriminant. Since all of the square roots were whole numbers, i could have used that as a reason why all the solutions would be rational, but its not really proving that all values for m will follow the rule. It just shows that 10 of my chosen values work.
from here i wasnt really sure what to do. I noted that the discriminant of the original equation was a quadratic function itself, so i graphed it, and noticed that for every integer value of x, its corrosponding y value will be a perfect square number. I wrote about this observation, and am just hoping its close enough to what i should have done.
If anybody has any idea about if i should have gone about this another way, or if i was right, please post :) all comments greatly appreciated.
Thanks,
Dan.
We were given the following quadratic equation:
12x^2 + 8mx + (4m-3) = 0
What we had to do was prove that for any integer value of m, the equation would have rational solutions.
What i did was first take the discriminant of the equation:
\Delta = (8m)^2 - 4(12)(4m-3)
simplified it:
\Delta = 64m^2 - 192m + 144
From that, i then created a table of values. I made a table of values for m (-5 to 5), the discriminant, then the square root of the discriminant. Since all of the square roots were whole numbers, i could have used that as a reason why all the solutions would be rational, but its not really proving that all values for m will follow the rule. It just shows that 10 of my chosen values work.
from here i wasnt really sure what to do. I noted that the discriminant of the original equation was a quadratic function itself, so i graphed it, and noticed that for every integer value of x, its corrosponding y value will be a perfect square number. I wrote about this observation, and am just hoping its close enough to what i should have done.
If anybody has any idea about if i should have gone about this another way, or if i was right, please post :) all comments greatly appreciated.
Thanks,
Dan.
