Solve :- 2x2 - (3+7i) x - (3-9i) = 0 ANURAG - Maths - Complex Numbers and Quadratic Equations

Question

Solve :- 2x2 - (3+7i) x - (3-9i) =
0
ANURAG

Harleen Birdi · Accepted Answer

We have 2x2 â€“ (3 + 7i) x â€“ (3 â€“ 9i) = 0 On comparing the given equation with the general form ax2 + bx + c = 0, we obtain a = 2, b = â€“ (3 + 7i), c = â€“ (3 â€“ 9i) Substituting the values of a, b, c in $α = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and β = \frac{- b \sqrt{a^{2} - 4 a c}}{2 a}, we get α= \frac{(3 + 7 i) + \sqrt{(3 + 7 i)^{2} + 9 (3 - 9 i)}}{4} and β = \frac{(3 + 7 i) - \sqrt{(3 + 7 i)^{2} + 8 (3 - 9 i)}}{4} \Rightarrow α = \frac{3 + 7 i + \sqrt{- 16 - 30 i}}{4} and β = \frac{3 + 7 i - \sqrt{- 16 - 30 i}}{4} (1)$ Now firstly find out $\sqrt{- 16 - 30 i}$ $Let a + i b = \sqrt{- 16 - 30 i} \Rightarrow a^{2} - b^{2} + 2 i a b = - 16 - 30 i \Rightarrow a^{2} - b^{2} = - 16 a n d 2 a b = - 30 N o w, c o n s i d e r (a^{2} + b^{2})^{2} = (a^{2} - b^{2})^{2} + 4 a^{2} b^{2} \Rightarrow (a^{2} + b^{2})^{2} = 256 + 900 = 1156 \Rightarrow a^{2} + b^{2} = 34 {Now, a^{2} - b^{2} = - 16 and a^{2} + b^{2} = 34}^{} \Rightarrow a^{2} = 9 and b^{2} = 25 \Rightarrow a = \pm 3 and b = \pm 5$ Since 2ab = â€“30, a and b must be of opposite signs $∴ a = 3 and b = - 5 or, a = - 3 and b = 5 \Rightarrow \sqrt{- 16 - 30 i} = 3 - 5 i or, - 3 + 5 i$ Substituting either of these values in (1), we get $α = \frac{(3 + 7 i) + (3 - 5 i)}{4} and, β = \frac{(3 + 7 i) - (3 - 5 i)}{4} \Rightarrow α = \frac{3}{2} + \frac{1}{2} i and, β = 3 i$ Hence the roots of the given equation are $\frac{3}{2} + \frac{1}{2} i and 3 i$

Solve :- 2x2 - (3+7i) x - (3-9i) = 0 ANURAG.

Solve :- 2x² - (3+7i) x - (3-9i) = 0

ANURAG.