Question

Quadratic Equations:Solve the following quadratic equation using the quadratic formula: formula attached, picture -y^2 - 3 = -3y

Answer 1

We have the following equation:

$-y^2-3=-3y$

by moving -3y to the left hand side, we have

$-y^2+3y-3=0$

which is in the form

$ay^2+by+c=0$

By comparing both equations, we can see that a=-1, b=3 and c=-3. By substituting these values in the quadratic formula, we get

$y=\frac{-3\pm\sqrt[]{3^2-4(-1)(-3)}}{2(-1)}$

which gives

$y=\frac{-3\pm\sqrt[]{9-12}}{-2}$

which leads to

$y=\frac{-3\pm\sqrt[]{-3}}{-2}$

but

$\begin{gathered} \sqrt[]{-3}=\sqrt{3}i \\ \text{where} \\ i=\sqrt[]{-1},\text{ the imaginary number} \end{gathered}$

Therefore, the solutions are:

$\begin{gathered} y_1=\frac{-3+\sqrt[]{3i}}{-2} \\ \text{and} \\ y_2=\frac{-3-\sqrt[]{3}i}{-2} \end{gathered}$