Correct option is D
Let z = 7 − 24 i . We are looking for z = a + b i , such that: ( a + b i ) 2 = 7 − 24 i Expanding the left-hand side: ( a + b i ) 2 = a 2 + 2 a b i + ( b i ) 2 = a 2 − b 2 + 2 a b i Equating real and imaginary parts: a 2 − b 2 = 7 (1) 2 a b = − 24 (2) From equation (2): a b = − 12 => b = − 12 a Substitute into equation (1): a 2 − ( − 12 a ) 2 = 7 => a 2 − 144 a 2 = 7 Multiply both sides by a 2 : a 4 − 144 = 7 a 2 => a 4 − 7 a 2 − 144 = 0 Let u = a 2 , then: u 2 − 7 u − 144 = 0 Solve using the quadratic formula: u = 7 ± 49 + 576 2 = 7 ± 625 2 = 7 ± 25 2 u = 16 or − 9 => a 2 = 16 => a = ± 4 Using a b = − 12 : If a = 4 , then b = − 12 4 = − 3 If a = − 4 , then b = − 12 − 4 = 3 Thus, the square roots of 7 − 24 i are: ± ( 4 − 3 i ) \begin{aligned}&\text{Let } z = 7 - 24i. \text{ We are looking for } \sqrt{z} = a + bi, \text{ such that:} \\&(a + bi)^2 = 7 - 24i \\[5pt]&\text{Expanding the left-hand side:} \\&(a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi \\[5pt]&\text{Equating real and imaginary parts:} \\&a^2 - b^2 = 7 \quad \text{(1)} \\&2ab = -24 \quad \text{(2)} \\[5pt]&\text{From equation (2):} \\&ab = -12 \Rightarrow b = \frac{-12}{a} \\[5pt]&\text{Substitute into equation (1):} \\&a^2 - \left( \frac{-12}{a} \right)^2 = 7 \Rightarrow a^2 - \frac{144}{a^2} = 7 \\[5pt]&\text{Multiply both sides by } a^2: \\&a^4 - 144 = 7a^2 \Rightarrow a^4 - 7a^2 - 144 = 0 \\[5pt]&\text{Let } u = a^2, \text{ then:} \\&u^2 - 7u - 144 = 0 \\[5pt]&\text{Solve using the quadratic formula:} \\&u = \frac{7 \pm \sqrt{49 + 576}}{2} = \frac{7 \pm \sqrt{625}}{2} = \frac{7 \pm 25}{2} \\[5pt]&u = 16 \text{ or } -9 \Rightarrow a^2 = 16 \Rightarrow a = \pm 4 \\[5pt]&\text{Using } ab = -12: \\&\text{If } a = 4, \text{ then } b = \frac{-12}{4} = -3 \\&\text{If } a = -4, \text{ then } b = \frac{-12}{-4} = 3 \\[5pt]&\text{Thus, the square roots of } 7 - 24i \text{ are:} \\&\boxed{\pm (4 - 3i)}\end{aligned} Let z = 7 − 24 i . We are looking for z = a + bi , such that: ( a + bi ) 2 = 7 − 24 i Expanding the left-hand side: ( a + bi ) 2 = a 2 + 2 abi + ( bi ) 2 = a 2 − b 2 + 2 abi Equating real and imaginary parts: a 2 − b 2 = 7 (1) 2 ab = − 24 (2) From equation (2): ab = − 12 => b = a − 12 Substitute into equation (1): a 2 − ( a − 12 ) 2 = 7 => a 2 − a 2 144 = 7 Multiply both sides by a 2 : a 4 − 144 = 7 a 2 => a 4 − 7 a 2 − 144 = 0 Let u = a 2 , then: u 2 − 7 u − 144 = 0 Solve using the quadratic formula: u = 2 7 ± 49 + 576 = 2 7 ± 625 = 2 7 ± 25 u = 16 or − 9 => a 2 = 16 => a = ± 4 Using ab = − 12 : If a = 4 , then b = 4 − 12 = − 3 If a = − 4 , then b = − 4 − 12 = 3 Thus, the square roots of 7 − 24 i are: ± ( 4 − 3 i )
