Question

In the following exercises, solve by using the Quadratic Formula.

$(v+1)(v-5)-4=0$

Short answer

The solutions of the given equation are$v=2\pm \sqrt{13}$.

Step-by-step solution

Step 1. Write the quadratic equation in standard form.

The given equation is:

$(v+1)(v-5)-4=0$

The given equation can be written as:

$\begin{array}{rcl}{v}^2-5v+v-5-4& =& 0\\ v^2-4v-9& =& 0\end{array}$

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a{x}^2+bx+c=0$is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation, $a=1,b=-4,c=-9$. Substitute these values in Quadratic Formula.

$v=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(-9)}}{2\left(1\right)}$

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl}v& =& \frac{4\pm \sqrt{16+36}}{2}\\ v& =& \frac{4\pm \sqrt{52}}{2}\\ v& =& \frac{4}{2}\pm \frac{2\sqrt{13}}{2}\\ v& =& 2\pm \sqrt{13}\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $v=2\pm \sqrt{13}$.