WRITE QUADRATIC EQUATION WITH GIVEN COMPLEX ROOTS
Complex numbers a + bi and a - bi are called complex conjugates.
If z = a + bi we write its conjugate as z* = a - bi.
(i) Quadratics with real coefficients are called real quadratics. This does not necessarily mean that its zeros are real.
(ii) If a quadratic equation has rational coefficients and an irrational root of the form c + d√n, then c - d√n is also a root. These roots are radical conjugates.
(iii) If a real quadratic equation has Δ < 0 and c + di is a complex root then c - di is also a root. These roots are complex conjugates.
To find quadratic equation with given complex roots, we will use the formula given below.
a(x2 - (Sum of the roots)x + Product of the roots) = 0
Find all quadratic equations with real coefficients and roots of:
Problem 1 :
3 ± i
Solution :
3 ± i
The roots are (3 + i) and (3 - i)
a(x2 - (Sum of the roots)x + Product of the roots) = 0
|
Sum of roots : = 3 + i + 3 - i = 6 |
Product of roots : = (3 + i) (3 - i) = 32 - i2 = 9+1 = 10 |
a(x2 - 6x + 10) = 0 where a ≠ 0
Problem 2 :
1 ± 3i
Solution :
1 ± 3i
The roots are 1 + 3i and 1 - 3i
|
Sum of roots : = 1 + 3i + 1 - 3i = 2 |
Product of roots : = (1 + 3i) (1 - 3i) = 12 - (3i)2 = 1 + 9 = 10 |
a(x2 - 2x + 10) = 0 where a ≠ 0
Problem 3 :
-2 ± 5i
Solution :
-2 ± 5i
Roots are (-2 + 5i) (-2 - 5i)
|
Sum of roots : = -2 + 5i - 2 - 5i = -4 |
Product of roots : = (-2 + 5i) (-2 - 5i) = 4 - (5i)2 = 4 + 25 = 29 |
a(x2 + 4x + 29) = 0 where a ≠ 0
Problem 4 :
√2 ± i
Solution :
√2 ± i
The roots are √2 + i and √2 - i.
|
Sum of roots : = √2 + i + √2 - i = 2√2 |
Product of roots : = (√2 + i) (√2 - i) = √22 - i2 = 2 + 1 = 3 |
a(x2 - 2√2x + 3) = 0 where a ≠ 0
Problem 5 :
2 ± √3
Solution :
2 ± √3
The roots are (2 + √3) (2 - √3)
|
Sum of roots : = 2 + √3 + 2 - √3 = 4 |
Product of roots : = (2 + √3) (2 - √3) = 22 - √32 = 4 - 3 = 1 |
a(x2 - 4x + 1) = 0 where a ≠ 0
Problem 6 :
0 and -2/3
Solution :
0 and -2/3
x = 0 and x = -2/3
a(x - 0) (x + 2/3) = 0
a(x) (3x +2) = 0
Multiplying 3 on each sides.
a[(x) (3x + 2)] = 0
a(3x2 + 2x) = 0 where a ≠ 0
Problem 7 :
± i√2
Solution :
± i√2
The roots are (+ i√2) (- i√2)
|
Sum of roots : = i√2 + (-i√2) = 0 |
Product of roots : = i√2 (-i√2) = -i2(√2)2 = 1(2) = 2 |
a(x2 + 2) = 0 where a ≠ 0
Problem 8 :
-6 ± i
Solution :
-6 ± i
The roots are (-6 + i) (-6 - i)
|
Sum of roots : = (-6 + i) + (-6 - i) = -12 |
Product of roots : = (-6 + i) (-6 - i) = 36 - i2 = 37 |
a(x2 + 12x + 37) = 0 where a ≠ 0
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