Chapter 3, Problem 11P

Use 5-digit arithmetic with chopping to determine the roots of the following equation with Eqs. (3.12) and (3.13)

$x^2-5000.002x+10$

Compute percent relative errors for your results.

To determine

To calculate: The roots of the equation $x^2-5000.002x+10$ with 5-digit arithmeticchopping by the use of the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ and $x=\frac{-2c}{b\pm \sqrt{b^2-4ac}}$. Also, find the percent relative error.

Answer to Problem 11P

Solution:

The roots of the equation $x^2-5000.002x+10$ by the use of the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ are $4999.9$ with relative percent error of $0.002\%$ and $0.05$ with relative percent error of $2400\%$.

And, the roots of the equation $x^2-5000.002x+10$ by the use of the formula $x=\frac{-2c}{b\pm \sqrt{b^2-4ac}}$ are $200$ with relative percent error of $96\%$ and $0.002$ with relative percent error of approximately $0\%$.

Explanation of Solution

Given:

The equation, $x^2-5000.002x+10$.

Formula used:

The quadratic formula for the equation $a{x}^2+bx+c=0$ is,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Alternative formula for roots when $b^2>>4ac$,

$x=\frac{-2c}{b\pm \sqrt{b^2-4ac}}$

Relative error formula:

$\text{Relative error = }\left|\frac{\text{true value}-\text{approximate value}}{\text{true value}}\right|$

Calculation:

Consider the following equation,

$x^2-5000.002x+10=0$

Here, $a=1,b=-5000.002\text{ and }c=10$.

Therefore, the roots of the equation from the quadratic formula is,

$\begin{array}{c}x=\frac{5000.002\pm \sqrt{{\left(5000.002\right)}^2-4\times 10}}{2}\\ =\frac{5000.002\pm \sqrt{{\left(5000.002\right)}^2-4\times 10}}{2}\\ =\frac{5000.002\pm 4999.998}{2}\end{array}$

Thus,

$\begin{array}{c}{x}_1=\frac{5000.002+4999.998}{2}\\ =\frac{10000}{2}\\ =5000\end{array}$

And,

$\begin{array}{c}{x}_2=\frac{5000.002-4999.998}{2}\\ =\frac{0.004}{2}\\ =0.002\end{array}$

Now, chop to 5 digits and find the root as below,

$\begin{array}{c}x=\frac{5000.0\pm \sqrt{{\left(5000.0\right)}^2-4\times 10}}{2}\\ =\frac{5000.0\pm \sqrt{25000000-4\times 10}}{2}\\ =\frac{5000.0\pm \sqrt{24999960}}{2}\\ =\frac{5000.0\pm 4999.9}{2}\end{array}$

Solve for two different roots:

$\begin{array}{c}{x}_1{}^{\prime }=\frac{5000.0+4999.9}{2}\\ =\frac{9999.9}{2}\\ =4999.9\end{array}$

And,

$\begin{array}{c}{x}_2{}^{\prime }=\frac{5000.0-4999.9}{2}\\ =\frac{0.1}{2}\\ =0.05\end{array}$

Hence, the relative percent error of the first root $4999.9$ is,

$\begin{array}{c}\text{Error}=\left|\frac{5000-4999.9}{5000}\right|\times 100\%\\ =\left|\frac{0.1}{5000}\right|\times 100\%\\ =0.002\%\end{array}$

And, the relative percent error of the second root $0.05$ is,

$\begin{array}{c}\text{Error}=\left|\frac{0.002-0.05}{0.002}\right|\times 100\%\\ =\left|\frac{-0.048}{0.002}\right|\times 100\%\\ =2400\%\end{array}$

Hence, the roots of the equations are $4999.9$ with relative percent error of $0.002\%$ and $0.05$ with relative percent error of $2400\%$.

Now, for the provided equation $b^2>>4ac$, Therefore, use alternative formula to find the roots by chopping to 5-digit,

$\begin{array}{c}\frac{-2c}{b\pm \sqrt{b^2-4ac}}=\frac{-2\left(10\right)}{-5000.0\pm \sqrt{{\left(5000.0\right)}^2-4\times 10}}\\ =\frac{-2\left(10\right)}{-5000.0\pm \sqrt{25000000-4\times 10}}\\ =\frac{-20}{-5000.0\pm \sqrt{24999960}}\\ =\frac{-20}{-5000.0\pm 4999.9}\end{array}$

Solve for two different roots,

$\begin{array}{c}{x}_1{}^{\prime \text{}\prime }=\frac{-20}{-5000.+4999.9}\\ =\frac{-20}{-0.1}\\ =200\end{array}$

And,

$\begin{array}{c}{x}_1{}^{\prime \text{}\prime }=\frac{-20}{-5000.0-4999.9}\\ =\frac{-20}{-9999.9}\\ =0.002\end{array}$

Therefore, the relative percent error of the first root $200$ is,

$\begin{array}{c}\text{ Error}=\left|\frac{5000-200}{5000}\right|\times 100\%\\ =\left|\frac{4800}{5000}\right|\times 100\%\\ =96\%\end{array}$

And, the relative percent error of the second root $0.00200002$ is,

$\begin{array}{c}\text{Error}=\left|\frac{0.002-0.002}{0.002}\right|\times 100\%\\ =\left|\frac{0}{0.002}\right|\times 100\%\\ =0\%\end{array}$

Hence, the roots of the equations are $200$ with relative percent error of $96\%$ and $0.002$ with relative percent error of $0\%$.