1. ## Quadratic Eqns. with Fractions

How do I solve the equation by completing the square and by the quadratic formula.
$\displaystyle 1/2x^2+2/3x+3/4=4/5$

2. Hello, HSS8!

How do I solve the equation by completing the square and by the quadratic formula?

. . $\displaystyle \frac{1}{2}x^2+\frac{2}{3}x+\frac{3}{4}\;=\;\frac{ 4}{5}$
Complete the square

We have: .$\displaystyle \frac{1}{2}x^2 + \frac{2}{3}x \:=\:\frac{1}{20}$

Multiply by 2: .$\displaystyle x^2 + \frac{4}{3}x \;=\;\frac{1}{10}$

Complete the square: .$\displaystyle x^2 + \frac{4}{3}x \:{\color{blue}+\:\frac{4}{9}} \;=\;\frac{1}{10}\:{\color{blue} \:+ \frac{4}{9}}$

Simplify: . $\displaystyle \left(x + \frac{2}{3}\right)^2 \;=\;\frac{49}{90}$

Square root: .$\displaystyle x + \frac{2}{3} \;=\;\pm\sqrt{\frac{49}{90}} \;=\;\pm\frac{7}{3\sqrt{10}} \;=\;\pm\frac{7\sqrt{10}}{30}$

Therefore: .$\displaystyle x \;=\;-\frac{2}{3}\pm\frac{7\sqrt{10}}{30} \quad\Rightarrow\quad \boxed{x \;=\;\frac{-20 \pm7\sqrt{10}}{30}}$

Quadratic Formula: .$\displaystyle x \;=\;\frac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have: .$\displaystyle \frac{1}{2}x^2 + \frac{2}{3}x + \frac{3}{4} \;=\;\frac{4}{5}$

Multiply by the LCD (60): .$\displaystyle 60\bigg[\frac{1}{2}x^2 + \frac{2}{3}x + \frac{3}{4}\;=\;\frac{4}{5}\bigg] \quad\Rightarrow\quad 30x^2 + 40x + 45 \;=\;48$

We have: .$\displaystyle 30x^2 + 40x - 3 \;=\;0\quad\Rightarrow\quad a = 30,\;b = 40,\;c = \text{-}3$

Then: .$\displaystyle x \;=\;\frac{-40 \pm\sqrt{40^2 - 4(30)(\text{-}3)}}{2(30)} \;=\;\frac{-40 \pm\sqrt{1960}}{60}$

. . . . $\displaystyle x \;=\;\frac{-40 \pm14\sqrt{10}}{60} \;=\;\boxed{\frac{-20 \pm 7\sqrt{10}}{30}}$

3. Thanks. When solving by completing the square, is it possible multiply by the LCD and then complete the square or will that lead to a wrong answer?

4. no change as long as every term is multiplied by the LCD, so go for it ... makes life easier for those suffering from phractional phobia, but you pay the price during any final simplification (reducing those large numbers).