Question

what are the roots of this quadratic equation?
$3x^2 + 10 = 4x$

a. $x = -4 \pm 6i\sqrt{26}$

b. $x = \frac{2 \pm i\sqrt{26}}{2}$

c. $x = \frac{-2 \pm i\sqrt{26}}{3}$

d. $x = \frac{2 \pm i\sqrt{26}}{3}$

Explanation:

Step1: Rewrite the equation in standard form

First, we rewrite the quadratic equation \(3x^{2}+10 = 4x\) into the standard form \(ax^{2}+bx + c=0\). Subtract \(4x\) from both sides:
\(3x^{2}-4x + 10=0\)
Here, \(a = 3\), \(b=-4\), and \(c = 10\).

Step2: Use the quadratic formula

The quadratic formula for the roots of a quadratic equation \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute \(a = 3\), \(b=-4\), and \(c = 10\) into the discriminant formula:
\(\Delta=(-4)^{2}-4\times3\times10=16 - 120=- 104\)
Since the discriminant is negative, we will have complex roots. We know that \(\sqrt{-104}=\sqrt{104}\times\sqrt{-1}=\sqrt{4\times26}\times i = 2\sqrt{26}i\) (where \(i=\sqrt{-1}\)).
Now, substitute \(a = 3\), \(b=-4\), and \(\Delta=- 104\) into the quadratic formula:
\(x=\frac{-(-4)\pm\sqrt{-104}}{2\times3}=\frac{4\pm2\sqrt{26}i}{6}\)
Simplify the fraction by dividing numerator and denominator by 2:
\(x=\frac{2\pm i\sqrt{26}}{3}\)

Answer:

D. \(x=\frac{2\pm i\sqrt{26}}{3}\)