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Question 608402: The coefficient of the second degree term of a quadratic equation is 1. If the constant term of the equation is increased by 1, the roots of the resulting equation are equal; but if it is diminished by 3, one root of the resulting equation is double the other. Find the original quadratic equation and its roots. So far, I've came up with ax^2+bx+(c-3) and ax^2+bx+(c-1). I don't know if I should be using the standard formula, vertex formula, root formula, etc. Please help me?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! You just have to read every sentence carefully to see what it is saying. We know that the leading coefficient is 1, so the polynomial is in the form
According to the second sentence, the polynomial ) has equal roots. This means that the discriminant is zero, i.e.
 = 0 \Rightarrow b^2 - 4c = 4)
According to the third sentence, the polynomial ) has two roots, one of which is double the other. Suppose we find the roots:
}}{2})
Therefore if we let -b = 3k and sqrt(b^2 - 4(c-3)) = k for some k, we will obtain two roots, one of which is double the other (since 3k + k = 2(3k - k)). Hence,
} = 3\sqrt{b^2 - 4c + 12})
We have a system of two equations but this is easily solvable since we know that b^2 - 4c = 4. Substitute this to obtain
 + 12} = 12 \Rightarrow b = -12)
Solving for c, we obtain c = 35. Therefore the original quadratic is
 and the roots are 5 and 7 (solved by factoring).
Update: I let -b = 3k and sqrt(b^2 - 4(c-3)) = k since that will produce two roots in the ratio 2:1. The roots of the quadratic now become (3k+k)/2 and (3k-k)/2, which are equal to 2k and k, in the ratio 2:1.
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