Chapter 7.2, Problem 3ST

To determine

To solve: The given equation by quadratic formula.

Answer to Problem 3ST

The solution of the equation is $w_{1,2}=\frac{5\pm \sqrt{37}}{2}$ .

Explanation of Solution

Given:

The equation is

  $w^2-5w=3$

Calculation:

The solution of the given equation is obtained by using the quadratic formula.

The given equation can be written as

  $w^2-5w-3=0$

For a quadratic equation in the form of $a{x}^2+bx+c=0$ , the solution of this equation is given by

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing the given equation with the standard quadratic equation, the coefficients are:

  $\begin{array}{l}a=1\\ b=-5\\ c=-3\end{array}$

Substituting these in the formula for the roots.

First take the positive sign.

  $\begin{array}{c}{w}_1=\frac{-\left(-5\right)+\sqrt{{\left(-5\right)}^2-4\times 1\times \left(-3\right)}}{2\times 1}\\ =\frac{5+\sqrt{25+12}}{2}\\ =\frac{5+\sqrt{37}}{2}\end{array}$

Take the negative sign.

  $\begin{array}{c}{w}_1=\frac{-\left(-5\right)-\sqrt{{\left(-5\right)}^2-4\times 1\times \left(-3\right)}}{2\times 1}\\ =\frac{5-\sqrt{25+12}}{2}\\ =\frac{5-\sqrt{37}}{2}\end{array}$

Therefore, the solution of the given radical equation is

  $\begin{array}{l}{w}_1=\frac{5+\sqrt{37}}{2}\\ w_2=\frac{5-\sqrt{37}}{2}\end{array}$ .

Therefore, the solution of the equation is $w_{1,2}=\frac{5\pm \sqrt{37}}{2}$ .

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