Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part II
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2-6 : Quadratic Equations - Part II

$9{w^2} - 6w = 101$

Show All Steps Hide All Steps

Start Solution

First, we need to get the quadratic equation in standard form. This is,

$9{w^2} - 6w - 101 = 0$ Show Step 2

Now we need to identify the values for the quadratic formula.

$a = 9\hspace{0.25in}b = - 6\hspace{0.25in}c = - 101$ Show Step 3

Plugging these into the quadratic formula gives,

\begin{align*}w = \frac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 9 \right)\left( { - 101} \right)} }}{{2\left( 9 \right)}} = \frac{{6 \pm \sqrt {3672} }}{{18}} & = \frac{{6 \pm \sqrt {\left( {36} \right)\left( {102} \right)} }}{{18}}\\ & = \frac{{6 \pm 6\sqrt {102} }}{{18}} = \frac{{1 \pm \sqrt {102} }}{3}\end{align*}

The two solutions to this equation are then : $\require{bbox} \bbox[2pt,border:1px solid black]{{w = \frac{1}{3} - \frac{{\sqrt {102} }}{3}\,\,{\mbox{and }}w = \frac{1}{3} + \frac{{\sqrt {102} }}{3}}}$ .