Question

integrated ii > chapter 4: solving quadratic equations with complex solutions >
solve $3x^2 + 87 = 30x$ using any method.
$x = \square$ and $x = \square$

Explanation:

Step1: Rewrite the equation

First, we rewrite the given quadratic equation \(3x^{2}+87 = 30x\) in standard form \(ax^{2}+bx + c=0\). Subtract \(30x\) from both sides:
\(3x^{2}-30x + 87=0\)
We can simplify this equation by dividing all terms by 3:
\(x^{2}-10x + 29 = 0\)

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(x^{2}-10x + 29 = 0\), we have \(a = 1\), \(b=- 10\), and \(c = 29\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute the values of \(a\), \(b\), and \(c\):
\(\Delta=(-10)^{2}-4\times1\times29=100 - 116=- 16\)
Now, substitute into the quadratic formula:
\(x=\frac{-(-10)\pm\sqrt{-16}}{2\times1}=\frac{10\pm4i}{2}\) (since \(\sqrt{-16}=\sqrt{16}\times\sqrt{-1} = 4i\))
Simplify the fraction:
\(x = 5\pm2i\)

Answer:

\(x = 5 + 2i\) and \(x = 5 - 2i\)