I am really iffy as to this problem.
One solution to a quadratic equation is “1-3i”. The coeffecients of the equations terms are rational. What is the constant term when the equation is in standard form.
The solution I got is x^2 + 2 x + 10 = 0.
Constant term is 10
That’s right, but for questions like these it’s always a good idea to post your logic so that if you don’t have the right answer we can help you to find your mistakes.
Your 10 is correct, but in your polynomial, the coefficient of x should be -2, not 2.
Thanks
Generally, if one solution to a quadratic with real terms is a + bi (with a and b real), the other solution is a - bi. So the two solutions must be 1 ± 3i, and the polynomial must be:
(x - 1 - 3i)(x - 1 + 3i)
To clarify this, Giles is using the mathematical definition of “generally” here. That’s not “It’s a pretty good bet that the other solution is probably a-bi”, that’s “In all cases and under all circumstances subject to the stated conditions, the other solution is always a-bi”.
Which is easily seen to be true: Since the equation has real coefficients, its complex conjugate is the same equation, and thus, if X is a solution, its complex conjugate must also be a solution.
…easily seen once you check that complex conjugation respects multiplication.
However, unless you specify that the leading coefficient is 1, there is no unizue solution. If the equation is x^2 - 2x +10 =0, then 2x^2 -4x - 20 has the same roots.
Except the problem does specify that the constant term is 10, so the second solution doesn’t fit the problem.
That the constant term is 10 is the answer to the problem, not a given piece of information.
However, it is given that the polynomial is to be in “standard form”, which I expect means that the leading coefficient is to be 1.