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=(-b\pm√(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=(36)/(8)\pm\frac{4\sqrt[]{6}i}{8} \\ x=(9)/(2)\pm\frac{\sqrt[]{6}i}{2} \end{gathered}`

Therefore, the solution in the simplest form is

`x=(9)/(2)\pm\frac{\sqrt[]{6}i}{2}`