Today, I realized that I am afraid of fraction calculation and decided to solve a quadratic equation o overcome this fear, step out of my comfort zone and practice my calculation skills.
Solve general form quadratic equations with fractions
x 2 + b a x + c a = 0 ( x + b 2 a ) 2 = ( b 2 a ) 2 − c a = b 2 4 a 2 − c a ⇒ x + b 2 a = ± b 2 − 4 a c 2 a ⇒ x = − b ± b 2 − 4 a c 2 a \begin{aligned}
&x^{2}+ \frac{b}{a}x+ \frac{c}{a}=0 \\
&(x+\frac{b}{2a})^{2}=(\frac{b}{2a})^{2}-\frac{c}{a}=\frac{b^{2}}{4a^{2}}-\frac{c}{a} \\
&\Rightarrow x+\frac{b}{2a} = \pm \frac{\sqrt{b^{2}-4ac}}{2a}\\
&\Rightarrow x =\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}
\end{aligned} x 2 + a b x + a c = 0 ( x + 2 a b ) 2 = ( 2 a b ) 2 − a c = 4 a 2 b 2 − a c ⇒ x + 2 a b = ± 2 a b 2 − 4 a c ⇒ x = 2 a − b ± b 2 − 4 a c y 2 − 11 6 y + 1 2 = 0 ( y − 11 6 ÷ 2 ) 2 = ( 11 6 ÷ 2 ) 2 − 1 2 = ( 11 12 ) 2 − 1 2 = 49 144 ⇒ y − 11 6 ÷ 2 = ± 49 144 = ± 7 12 ⇒ y − 11 2 = ± 7 12 ⇒ y = 1 3 o r 3 2 \begin{aligned}
&&y^{2}-\frac{11}{6}y+\frac{1}{2}&=0\\
&&(y-\frac{11}{6}\div2)^{2}&=(\frac{11}{6}\div2)^{2}-\frac{1}{2}\\
&& &=(\frac{11}{12})^{2}-\frac{1}{2}\\
&& &=\frac{49}{144}\\
& \Rightarrow &y-\frac{11}{6}\div2&= \pm \sqrt{\frac{49}{144}}\\
&& &=\pm\frac{7}{12}\\
& \Rightarrow &y-\frac{11}{2}&=\pm\frac{7}{12}\\
& \Rightarrow &y&=\frac{1}{3} or \frac{3}{2}
\end{aligned} ⇒ ⇒ ⇒ y 2 − 6 11 y + 2 1 ( y − 6 11 ÷ 2 ) 2 y − 6 11 ÷ 2 y − 2 11 y = 0 = ( 6 11 ÷ 2 ) 2 − 2 1 = ( 12 11 ) 2 − 2 1 = 144 49 = ± 144 49 = ± 12 7 = ± 12 7 = 3 1 or 2 3