Question

solve the following equation by either factoring or quadratic formula $5x^2 - 11x = 0$
enter your answer as a decimal
$x_1 = \square$
$x_2 = \square$

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation \(5x^{2}-11x = 6\) in standard quadratic form \(ax^{2}+bx + c=0\). Subtract 6 from both sides:
\(5x^{2}-11x - 6=0\)
Here, \(a = 5\), \(b=- 11\), \(c=-6\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\).
Substitute \(a = 5\), \(b=-11\), \(c = - 6\) into the discriminant formula:
\(\Delta=(-11)^{2}-4\times5\times(-6)=121 + 120=241\)
Then, find the roots using the quadratic formula:
\(x=\frac{-(-11)\pm\sqrt{241}}{2\times5}=\frac{11\pm\sqrt{241}}{10}\)
Calculate the numerical values:
\(\sqrt{241}\approx15.524\)
For the plus sign: \(x_1=\frac{11 + 15.524}{10}=\frac{26.524}{10}=2.6524\approx2.65\) (or more accurately, let's recalculate \(\sqrt{241}\approx15.524174696\), so \(x_1=\frac{11 + 15.524174696}{10}=\frac{26.524174696}{10}=2.6524174696\approx2.65\) (or we can also factor the equation. Let's try factoring:
We need two numbers that multiply to \(5\times(-6)=-30\) and add up to \(-11\). The numbers are \(-15\) and \(2\).
Rewrite the middle term: \(5x^{2}-15x + 2x-6 = 0\)
Factor by grouping: \(5x(x - 3)+2(x - 3)=0\)
\((5x + 2)(x - 3)=0\)
Set each factor equal to zero:
\(5x+2 = 0\) or \(x - 3=0\)
For \(5x+2 = 0\), \(5x=-2\), \(x=-\frac{2}{5}=-0.4\)
For \(x - 3=0\), \(x = 3\)
So the roots are \(x_1 = 3\) and \(x_2=-0.4\) (or \(x_1=-0.4\) and \(x_2 = 3\))

Answer:

\(x_1 = 3.0\), \(x_2=-0.4\) (or \(x_1=-0.4\), \(x_2 = 3.0\))