Question

Q1.Find the positive root of the equation x^3 + 2x^2 + 10x – 20 using Regula Falsi method and correct upto 4 decimal places.
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Answer 1

Step-by-step explanation:

Given:x³+2x²+10x-20=0

To find: find the real root correct to three significant figure by regula falsi method.

Solution:

Formula:

$\begin{gathered}\boxed{\bold{x = \frac{bf(a) - af(b)}{f(a) - f(b)}} } \\ \end{gathered}$

Step 1:Find a and b,put x=1

$\begin{gathered}f(1) = {(1)}^{3} + 2( {1)}^{2} + 10(1) - 20 \\ \\ f(1) = 1 + 2 + 10 - 20 \\ \\ \bold{\red{f(1) = - 7}} \\ \\ \end{gathered}$

Thus, a=1 and f(a)= -7

put x=2

$\begin{gathered}f(2) = {(2)}^{3} + 2( {2)}^{2} + 10(2) - 20 \\ \\ f(2) = 8 + 8 + 20 - 20 \\ \\ \bold{\green{f(2) = 16}} \\ \\ \end{gathered}$

Thus, b=2 and f(b)=16

There is a real root between x=1 and 2.

Step 2: Find next a,by putting value in formula

$\begin{gathered} x_1 = \frac{2( - 7) - 1(16)}{( - 7) - (16)} \\ \\ x_1 = \frac{ - 14 - 16}{ - 7 - 16} \\ \\ x_1 = \frac{ - 30}{ - 23} \\ \\ \bold{\pink{x_1 = 1.3043}}\\ \end{gathered}$

find the value of f(1.3043)

$\begin{gathered}f(1.3043) = {(1.3043)}^{3} + 2( {1.3043)}^{2} + 10(1.3043) - 20 \\ \\ \bold{\pink{f(1.3043) = - 1.3357}} \\ \\\end{gathered}$

Step 3:For next iteration a=1.3043 and f(a)=-1.3357

$\begin{gathered}x_2 = \frac{2( - 1.3357) - 1.3043(16)}{ - 1.3357- (16)} \\ \\ x_2 = \frac{ - 2.6714- 20.8688}{ - 17.3357} \\ \\ x_2 = \frac{-23.5402}{ - 17.3357} \\ \\ \bold{\green{x_2 = 1.3579}}\\\end{gathered}$

$\begin{gathered}f(1.3579) = {(1.3579)}^{3} + 2( {1.3579)}^{2} + 10(1.3579) - 20 \\ \\ f(1.3579) = 2.5038 + 3.6877 + 13.579 - 20 \\ \\ \bold{\green{f(1.3579) = - 0.2295}} \\ \\\end{gathered}$

Step 4: For next iteration a=1.3579 and f(a)=-0.2295

$\begin{gathered}x_3= \frac{2( - 0.2295) - 1.3579(16)}{ - 0.2295 - 16} \\ \\ x_3 = \frac{ - 22.1854}{ - 16.2295} \\ \\ \bold{\purple{x_3 = 1.366}} \\ \end{gathered}$

Find f(1.366)

$\begin{gathered}f(1.3669) = {(1.3669)}^{3} + 2( {1.3669)}^{2} + 10(1.3669) - 20 \\ \\ \bold{\purple{f(1.3669) = - 0.0402}} \\ \end{gathered}$

Step 5: For next iteration a=1.3669 and f(a)=-0.0402

$\begin{gathered}x_4= \frac{2( - 0.0402) - 1.3669(16)}{ - 0.0402 - 16} \\ \\ x_4 = \frac{ - 21.9508}{ - 16.0402} \\ \\ \bold{\red{x_4= 1.3684}} \\ \end{gathered}$

*These steps can be iterate more to find more accurate root.

The real root exists more accurately at 1.3688

Final answer:

Real root of x³+2x²+10x-20=0 is approx 1.3688 upto 4 significant digits.

Hope it helps you.

To learn more on brainly:

Find the real root of the equation x³ -9x+1=0 by using regula falsi method.

https://brainly.in/question/2930619