Question

use the quadratic formula to solve the quadratic equation $10x^2 + 5x = 5x - 8$ express its solutions in the form $a \pm bi$. (1 point)

Explanation:

Step1: Simplify the equation

First, we need to rewrite the given quadratic equation \(10x^{2}+5x = 5x - 8\) in standard form \(ax^{2}+bx + c=0\). Subtract \(5x\) from both sides and add \(8\) to both sides:
\(10x^{2}+5x-5x + 8=5x - 8-5x + 8\)
Simplifying, we get \(10x^{2}+8 = 0\). So, \(a = 10\), \(b=0\), \(c = 8\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 10\), \(b = 0\), \(c=8\) into the formula:
\(x=\frac{-0\pm\sqrt{0^{2}-4\times10\times8}}{2\times10}\)
First, calculate the discriminant \(\Delta=b^{2}-4ac=0 - 320=- 320\)
Then, \(\sqrt{\Delta}=\sqrt{-320}=\sqrt{320}\times\sqrt{- 1}=\sqrt{64\times5}\times i = 8\sqrt{5}i\) (since \(\sqrt{-1}=i\))
So, \(x=\frac{\pm8\sqrt{5}i}{20}\)
Simplify the fraction: \(x=\frac{\pm2\sqrt{5}i}{5}=0\pm\frac{2\sqrt{5}}{5}i\)

Answer:

\(0\pm\frac{2\sqrt{5}}{5}i\)