Question

Solve -x2 - 3x = 4 over the set of complex numbers

Answer 1

Answer:

$\bold{x=\dfrac{-3\pm i\sqrt{7}}{2}}$

Step-by-step explanation:

Given quadratic equation is:

$-x^2 - 3x = 4$

Rewriting the given equation:

$-x^2 - 3x - 4 = 0$

OR

$x^2 + 3x + 4=0$

Solution of a quadratic equation represented as $ax^2+bx+c=0$ is given as:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Comparing the given equation with standard equation:

a = 1

b = 3

c = 4

So, the roots are:

$x=\dfrac{-3\pm\sqrt{3^2-4\times 1 \times 4}}{2\times 1}\\\Rightarrow x=\dfrac{-3\pm\sqrt{9-16}}{2}\\\Rightarrow x=\dfrac{-3\pm\sqrt{-7}}{2}$

$\sqrt{-7 }$ can be written as $7\sqrt{-1}$

and $\sqrt{-1} = i$

So, $\sqrt{-7} = i\sqrt7$

The numbers containing $i$ in them, are called as complex numbers.

Therefore, the roots of the equation can be written as:

$\Rightarrow \bold{x=\dfrac{-3\pm i\sqrt{7}}{2}}$