solve x2-(2-3i)x+(10+4i)=0

Dear student using roots for quadratic equationx=(2-3i)±(2-3i)2-4(10+4i)2=(2-3i)±4+9i2-12i-40-16i2=(2-3i)±-28i-452=(2-3i)±-49+4-28i2=(2-3i)±(7i-2)22=2-3i+7i-22,2-3i-7i+22=2i,2-5i Regards

solve x²-(2-3i)x+(10+4i)=0

Dear student

u \sin g r o o t s f o r q u a d r a t i c e q u a t i o n x = \frac{(2 - 3 i) \pm \sqrt{(2 - 3 i)^{2} - 4 (10 + 4 i)}}{2} = \frac{(2 - 3 i) \pm \sqrt{4 + 9 i^{2} - 12 i - 40 - 16 i}}{2} = \frac{(2 - 3 i) \pm \sqrt{- 28 i - 45}}{2} = \frac{(2 - 3 i) \pm \sqrt{- 49 + 4 - 28 i}}{2} = \frac{(2 - 3 i) \pm \sqrt{(7 i - 2)^{2}}}{2} = \frac{2 - 3 i + 7 i - 2}{2}, \frac{2 - 3 i - 7 i + 2}{2} = 2 i, 2 - 5 i

Regards

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