Answer:
Below
Explanation:
Question 1:
The quadratic equation is of the form ax^2 + bx + c = 0, where a, b, and c are constants. If one solution of the equation is 7-13i, the other solution must be the complex conjugate of 7-13i, which is 7+13i. To find the complex conjugate of a complex number, you simply need to change the sign of the imaginary part of the number. In this case, the complex conjugate of 7-13i is 7+13i.
Question 2:
1. (4-3i) (-7+21)
To add two complex numbers, we need to add their real parts and imaginary parts separately. In this case, we have:
(4-3i) (-7+21)
= (4 - 7) + (-3i + 21)
= -3 + 18i
2. (-7+6i)+(2-3i)+(4+5i)
To add multiple complex numbers, we can add them one at a time using the same procedure as before. In this case, we have:
(-7+6i)+(2-3i)+(4+5i)
= (-7+2) + (6i - 3i) + (4 + 5i)
= -5 + 3i + 4 + 5i
= -1 + 8i
3. (-6-i)/4-2i
To divide two complex numbers, we need to multiply the numerator and denominator by the complex conjugate of the denominator. In this case, we have:
4. (-6-i)/4-2i
= (-6-i)(4+2i)/(4-2i)(4+2i)
= (-24 - 2i + 6i - i^2)/20
= (-24 + 4 -1)/20
= (-21)/20
= -1.05 - 0.05i
5. (-2-7i)(-2+7i)
To multiply two complex numbers, we use the FOIL method, which involves multiplying the First, Outer, Inner, and Last terms separately and then adding the results. In this case, we have:
(-2-7i)(-2+7i)
= (-2 * -2) + (-2 * 7i) + (-7i * -2) + (-7i * 7i)
= 4 + (-14i) + 14i + 49
= 4 + 49
= 53
6. (5-3i)-(-2-6i)
To subtract two complex numbers, we need to add the first number to the opposite of the second number. In this case, we have:
(5-3i)-(-2-6i)
= (5-3i) + (2 + 6i)
= 7 + 3i
This is the final answer for each of the problems.