osteoblogda

2021-12-29

To solve: $\frac{{x}^{2}}{2}+\frac{x}{20}=\frac{1}{10}$

einfachmoipf

Calculation:
Multiply by 20 on both sides,
$20\left(\frac{{x}^{2}}{2}+\frac{x}{20}\right)=\frac{1}{10}\cdot 20$
By distributive property,
$20\left(\frac{{x}^{2}}{2}+\frac{x}{20}\right)=20\cdot \frac{{x}^{2}}{2}+20\cdot \frac{x}{20}$
$20\left(\frac{{x}^{2}}{2}+\frac{x}{20}\right)=10{x}^{2}+x$
$\frac{1}{10}\cdot 20=2$
$10{x}^{2}+x=2$
Subtract 2 on both sides,
$10{x}^{2}+x-2=2-2$
$10{x}^{2}+x-2=0$
Split the middle term .
Replace
$10{x}^{2}-4x+5x-2=0$
The above equation can be written as $2x\cdot 5x-2\cdot 2x+1\cdot 5x-1\cdot 2$
Apply distributive property for first two terms and last two terms,
$2x\cdot 5x-2\cdot 2x+1\cdot 5x-1\cdot 2=2x\left(5x-2\right)+1\left(5x-2\right)$
Again apply distributive property for $2x\left(5x-2\right)+1\left(5x-2\right)$
$2x\left(5x-2\right)+1\left(5x-2\right)=\left(2x+1\right)\left(5x-2\right)$
So the factors are
$10{x}^{2}+x-2=\left(2x+1\right)\left(5x-2\right)$
To find the value of x,
Take $5x-2=0$
$5x-2+2=0+2$
$5x=2$
Divide by 5 on both sides,
$\frac{5x}{5}=\frac{2}{5}$
$x=\frac{2}{5}$
Take $2x+1=0$
Subtract 1 on both sides,
$2x+1-1=0-1$
$2x=-1$
Divide by 2 on both sides,

Paul Mitchell

Step 1: Simplify both sides of the equation.
$\frac{1}{2}{x}^{2}+\frac{1}{20}x=\frac{1}{10}$
Step 2: Subtract 1/10 from both sides.
$\frac{1}{2}{x}^{2}+\frac{1}{20}x-\frac{1}{10}=\frac{1}{10}-\frac{1}{10}$
$\frac{1}{2}{x}^{2}+\frac{1}{20}x+\frac{-1}{10}=0$
For this equation: $a=0.5,b=0.05,c=-0.1$
$0.5{x}^{2}+0.05x±0.1=0$
Step 3: Use quadratic formula with $a=0.5,b=0.05,c=-0.1$.
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-\left(0.05\right)±\sqrt{{\left(0.05\right)}^{2}-4\left(0.5\right)\left(-0.1\right)}}{2\left(0.5\right)}$
$x=\frac{-0.05±\sqrt{0.2025}}{1}$
$x=0.4,-0.5$
Answer: $x=0.4,-0.5$

karton

$\begin{array}{}\frac{1}{2}{x}^{2}+\frac{1}{20}x=\frac{1}{10}\\ \frac{1}{2}{x}^{2}+\frac{1}{20}x-\frac{1}{10}=\frac{1}{10}-\frac{1}{10}\\ \frac{1}{2}{x}^{2}+\frac{1}{20}x-\frac{1}{10}=0\\ x=\frac{-\frac{1}{20}±\sqrt{\left(\frac{1}{20}{\right)}^{2}-4×\left(\frac{1}{2}\right)\left(-\frac{1}{10}\right)}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\sqrt{\frac{1}{400}-4×\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\sqrt{\frac{1}{400}-2\left(-\frac{1}{10}\right)}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\sqrt{\frac{1}{400}+\frac{1}{5}}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\sqrt{\frac{81}{400}}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\frac{9}{20}}{2×\left(\frac{1}{2}\right)}\\ x=\frac{-\frac{1}{20}±\frac{9}{20}}{1}\\ x=\frac{\frac{2}{5}}{1}\\ x=\frac{2}{5}\\ x=\frac{-\frac{1}{2}}{1}\\ x=-\frac{1}{2}\\ x=\frac{2}{5}\\ x=-\frac{1}{2}\end{array}$