Is it possible for a quadratic function to have one real root and one imaginary root? Justify your answer.?

Question

Is it possible for a quadratid function to have only one real root? Justify your answer.

Is it possible for a quadratic function to have only one imaginary root? Justify your answer.

Answer 1

Only if the it's coefficients are not all real numbers. Otherwise, the answer for both questions is no.

If f is a quadratic function, that is, a polynomial of degree 2 with real coefficients, then

f(x) = ax^2 + bx + c = a<>0. As we know, the roots of f are given by Bhaskara formula

x = (- b + or - sqrt(b^2 - 4ac))/2a.

If b^2 - 4ac >=0, then sqrt(b^2 - 4ac) is areal number and, since a, b and c are real, both roots (which can be equal) are real. And if b^2 - 4ac <0, then sqrt(b^2 - 4ac) is imaginary and both roots are complex (not real).

So, a, b and c are real, the conditions mentioned cannot happen.

But if you allow at least one of the coefficients a, b and c be non real complex, then the answer is yes. For example, the roots of the polynomial

p(x) = (x-1) (x -i) = x^2 -(1 + i)x + i are 1 and i, one real, other pure imaginary.

Answer 2

When quadratics have imaginary roots they come in pairs from taking both the positive and negative square root of a negative number. Since quadratics have at most two roots they are either real or imaginary, not both.

It is possible to have one real root. For example

x^-2x+1 = 0 factors to (x-1)^2=0 has one root x = 1.

It is not possible for a quadratic equation to have just one root imaginary root. Imaginary roots come from taking both the positive and the negative square root of a negative real number such as 3 +/- sqrt(-4) = 3 +/- 2i

Answer 3

The quadratic formula is:

-b ± sqrt(b² - 4ac)
------------------------
........... 2a

a, b and c are all real numbers.

The only thing that can make an imaginary number is the sqrt of a negative number. But then you will have two imaginary roots (± sign). The only choices are:
1 real root, 2 real roots, 2 imaginary roots.

Answer 4

No it can only have two real roots or two imaginary roots. Only polynomials of odd orders >= 3 can have both imaginary and real roots.

Example

(x+1)(x^2+x+1) -> has 1 real root and two imaginary roots.

Answer 5

The answer is YES. Justification:

Let

y = (ax+b)(cx+id) = acx^2 + x(bc + iad) + ibd

with a,b,c,d real and i = sqrt(-1); this is the general quadratic form with one real and one imaginary root, by definition, since I have started with factors that are zero for x=-b/a and x=-id/c.

Answer 6

No.
b^2-4ac only yields one result. If it is negative, then the roots are imaginary.

Answer 7

I think you should go to yahoo and ask for homework help that helped me.

Answer 8

I'm not sure, I spose so.