Question

Write the quadratic equation whose roots are 2+3i and 2-3i

Answer 1

f(x) = [x - (2 + 3i)] [x - (2 - 3i)] = x^2 - (2 + 3i)x - (2 - 3i)x + (2 + 3i)(2 - 3i)

f(x) = x^2 - 2x -3i x - 2x + 3i x + (4 - 9i^2)

f(x) = x^2 - 4x + 13

I hope it will help you

Answer 2

Given,

• The roots of the quadratic equation are 2+3i and 2-3i.

To find,

• We have to find the quadratic equation whose roots are 2+3i and 2-3i.

Solution,

We can simply find the quadratic equation whose roots are 2+3i and 2-3i  by using the following formula:

    x² - ( sum of the roots )x + (product of the roots) = 0      (*)

The roots of the quadratic equation are 2+3i and 2-3i.

Sum of the roots = 2+3i + 2-3i

Sum of the roots = 4

Product of the roots = (2+3i)(2-3i)

Using (a+b)(a-b) = a²-b², we get

     a = 2, b = 3i

then, product of the roots = 4 - (3i)²

                                           = 4 + 9

                                           = 13

Product of the roots = 7

Using (*), we get

   x²- (4)x + 13

    x²-4x+13

Hence, the quadratic equation whose roots are 2+3i and 2-3i is x²-4x+13.