Write a quadratic equation with the given roots: 2 and 1 - i?

Question

please show work!

Answer 1

The only way this can be a quadratic equation is if it has complex coefficients. Just multiply out (x-2)(x-1+i) and gather like terms.

If it has real coefficients, then it cannot be a quadratic, but rather the lowest possible degree is cubic. If 1-i is a root, then 1+i is also a root (the conjugate roots theorem tells you this).

It's easy to build a factor from the given root of 2, it is (x-2). It's a little more of a challenge to build the factor from the pair of roots 1±i. Here's a little "recipe" to do that:

x² - (sum)x + product
where sum and product refer to the sum and product of the pair of conjugate roots. So, 1+i+1-i = 2 and (1+i)(1-i) = 1-i² = 2. Therefore, the factor is (x² - 2x + 2). Multiply this factor by (x-2) from above to get the resulting cubic:

x^3 -4x² - 2x - 4 = 0

Answer 2

Since the roots are 1/2 and 4/3, (x - 1/2)(x - 4/3) = 0 Expanding, x^2 - (1/2)x - (4/3)x + 2/3 = 0 x^2 - (11/6)x + 2/3 = 0 Multiplying both sides by 6, 6x^2 - 11x + 4 = 0 --- this is your quadratic equation. CHECK: Using the quadratic formula and noting that a = 6 b = -11 c = 4 the roots will come out to be x = 1/2 and x = 4/3 (which satisfy the given in the problem).

Answer 3

If you are asking for an equation with real coefficients there is no quadratic, as the complex roots appears by pares(conjugates) and then there are three roots
With complex coefficients the equation would be
(x-2)(x-1+i) = x^2-x+ix-2x+2+2i= x^2+x(-3+i)+2+2i=0

Answer 4

First, the quadratic will have complex coefficients,
because if the coefficients are real, the complex
conjugate must also be a root.
Next, if your quadratic equation is x²+px+q = 0,
the sum of its roots is -p and the product is q.
So your equation is
x² + (-1+i)x + (2-2i) = 0.

Answer 5

you have to use the factor theorem to write then factored form. to do this, write the binomial that, when set equal to zero, equals the roots. all you have to do is put an x inside parentheses and then use the opposite sign. so, if 2 is a root, (x - 2) is a factor. complex roots, such as 1 - i, always come in pairs, so 1 + i must also be a root. putting this all together, the factored form is as follows:

(x - 2)(x - (1 - i))(x + (1 + i))

the complex root needs to be in parentheses b/c you have to distribute the negative if there is one. to get the final answer, FOIL two of the binomials, multiply by the third, and combine like terms.

Answer 6

(x-2)[(x-(1-i)]=0
x^2-2x+2(1-i)=0