If (3 + 4i) ia a root of the equation

Question

x^2 + px + q then p = ? and q = ?

Answer 1

The equation is x² + px + q = 0.

The real number coefficients is 3 + 4i then the other root has to be 3 - 4i because the coefficient of x is minus the sum of the roots and the independent term is the product of the roots and they would have imaginary parts.

So the roots of the equation are x = 3 + 4iand x = 3 - 4i.

• If (3 + 4i) is a root of x² + px + q, then f(3 + 4i) = 0.

f(3+ 4i) = (3 + 4i)² + p(3 + 4i) + q = 0

(3 + 4i)² + p(3 + 4i) + q = 0 → ( 1 )

• If (3 - 4i) is a root of x² + px + q, then f(3 - 4i) = 0.

f(3- 4i) = (3 - 4i)² + p(3 - 4i) + q = 0

(3 - 4i)² + p(3 - 4i) + q = 0 → ( 2 )

Solve eqn (1) and (2) for p and q.

Write the equations in column form, then subtract them, and solve for p.

(3 + 4i)² + p(3 + 4i) + q = 0

(3 - 4i)² + p(3 - 4i) + q = 0

( - )____________________________

[ (3 + 4i)²- (3 - 4i)²] + p[ 3 + 4i - 3 + 4i ]= 0

[ (3 + 4i)²- (3 - 4i)²] + 8ip= 0

8ip = (3 - 4i)² - (3 + 4i)²

8ip = 9 - 24i + 16i² - (9 + 24i + 16i²)

8ip = 9 - 24i + 16i² - 9 - 24i - 16i²

8ip = - 24i - 24i = - 48i

8p = - 48

⇒ p = - 48/8 = - 6.

Substitute the value p = - 6 in eq (1), and solvefor q.

(3 + 4i)² - 6(3 + 4i) + q = 0

9 + 24i + 16i² - 18 - 24i + q = 0

- 9 + 16i² + q = 0

Substitute i² = -1.

- 9 - 16 + q = 0

⇒ q = 25.

The value of p = - 6 and q = 25.