Question

question
what are the roots of the equation $9x^2 = -48x - 67$ in simplest $a + bi$ form?
answer
attempt 1 out of 2
+ additional solution - no solution
$x = $

Explanation:

Step1: Rewrite the equation in standard form

First, we rewrite the given equation \(9x^{2}=-48x - 67\) in the standard quadratic form \(ax^{2}+bx + c = 0\). We add \(48x\) and \(67\) to both sides of the equation:
\(9x^{2}+48x + 67=0\)
Here, \(a = 9\), \(b = 48\), and \(c = 67\).

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is given by:
\(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
First, we calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute \(a = 9\), \(b = 48\), and \(c = 67\) into the discriminant formula:
\(\Delta=(48)^{2}-4\times9\times67\)
\(=2304-4\times9\times67\)
\(=2304 - 2412\)
\(=- 108\)

Now, substitute \(a = 9\), \(b = 48\), and \(\Delta=-108\) into the quadratic formula:
\(x=\frac{-48\pm\sqrt{- 108}}{2\times9}\)
We know that \(\sqrt{-108}=\sqrt{108}\times\sqrt{-1}=\sqrt{36\times3}\times i = 6\sqrt{3}i\) (since \(\sqrt{-1}=i\))

So,
\(x=\frac{-48\pm6\sqrt{3}i}{18}\)
We can simplify this fraction by dividing the numerator and the denominator by \(6\):
\(x=\frac{-8\pm\sqrt{3}i}{3}\)
Which can be written as \(x =-\frac{8}{3}\pm\frac{\sqrt{3}}{3}i\)

Answer:

The roots of the equation are \(x =-\frac{8}{3}+\frac{\sqrt{3}}{3}i\) and \(x =-\frac{8}{3}-\frac{\sqrt{3}}{3}i\)