Cubic Polynomial with Repeated Real Root
One of the roots of the following cubic polynomial is 2. The other root is also real and repeated. Solve for the repeated root.
x 3 − 1 2 x 2 + 4 5 x − 5 0 = 0
The answer is 5.
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2 solutions
This problem is solved by synthetic division. The problem said that one of the roots was 2.
x − 2 x 3 − 1 2 x 2 + 4 5 x − 5 0 = x 2 − 1 0 x + 2 5
The resulting quadratic polynomial can be solved using the Quadratic formula:
x = 2 a − b + / − b 2 − 4 a c = 2 1 0 + / − 1 0 2 − 4 ∗ 1 ∗ 2 5 = 5
If a is the double root then we have, by Viete, 2 + 2 a = 1 2 (negative the coefficient of x 2 ) so a = 5