Question

1. Solve the following quadratic equation x2 + 3ix + 10 = 0.​

Answer 1

Step-by-step explanation:

${x}^{2} + 3x + 10 = 0 \\ {x}^{2} + 5x - 2x + 10 = 0 \\ x(x + 5) - 2(x + 5) \\ (x + 5)(x - 2)$

I hope it helps you

Answer 2

Answer :

$\large \sf \implies x = \dfrac{-52i}{2} ,\dfrac{46i}{2}$

Solution :

Question :

• Solve the following quadratic equation x² + 3ix + 10 = 0.

Let's solve,

$\red\bigstar \: \boxed{x = \dfrac{-b±\sqrt{b^2 -4ac}}{2a}}$

Here,

• a = 1
• b = 3i
• c = 10

$\LARGE\color{green} \underline{\underline{\mathfrak{Substituting \: the\: values}}}$

$\large \sf x = \dfrac{-(3i) ±\sqrt{(3i)^2 -4(1)(10)}}{2a}$

$\large \sf x = \dfrac{-(3i) ±\sqrt{9i^2 -4(1)(10)}}{2a}$

$\:\:\:\:\;\:\:\:\:\red\bigstar\:\color{blue}\boxed{\sf i^2 = -1}$

$\large \sf \implies x = \dfrac{-3i ±\sqrt{9(-1)-40}}{2(1)}$

$\large \sf \implies x = \dfrac{-3i ±\sqrt{-9-40}}{2}$

$\large \sf \implies x = \dfrac{-3i ±\sqrt{-49}}{2}$

$\:\:\:\:\;\:\:\:\:\red\bigstar\:\color{blue}\boxed{ \sf i = \sqrt{-1}}$

So,

$\large \sf \implies x = \dfrac{-3i ± (-49i)}{2}$

$\large \sf \implies x = \dfrac{-3i + (-49i)}{2} ,\dfrac{-3i - (-49i)}{2}$

$\large \sf \implies x = \dfrac{-3i - 49i}{2} ,\dfrac{-3i +49i}{2}$

$\large \sf \implies x = \dfrac{-52i}{2} ,\dfrac{46i}{2}$

Hence,

$\large \sf \implies x = \dfrac{-52i}{2} ,\dfrac{46i}{2}$

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