Question

What are the roots of the equation 4x? = 36x - 87 in simplest a + bi form?

Answer 1

We are given the following quadratic equation

$4x^2=36x-87$

Let us first re-write the equation in the standard form as

$\begin{gathered} 4x^2=36x-87 \\ 4x^2-36x+87=0 \end{gathered}$

Recall that the standard form of a quadratic equation is given by

$ax^2+bx+c=0$

Comparing the given quadratic equation with the standard form, we see that the coefficients are

a = 4

b = -36

c = 87

Recall that the quadratic formula is given by

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

Let us substitute the values of the coefficients into the above quadratic formula.

$\begin{gathered} x=\frac{-(-36)\pm\sqrt[]{(-36)^2-4(4)(87)}}{2(4)} \\ x=\frac{36\pm\sqrt[]{1296-1392}}{8} \\ x=\frac{36\pm\sqrt[]{-96}}{8} \\ x=\frac{36\pm4\sqrt[]{6}i}{8} \\ x=\frac{36}{8}\pm\frac{4\sqrt[]{6}i}{8} \\ x=\frac{9}{2}\pm\frac{\sqrt[]{6}i}{2} \end{gathered}$

Therefore, the solution in the simplest form is

$x=\frac{9}{2}\pm\frac{\sqrt[]{6}i}{2}$