Question

Use the method of false position to find the fourth root of 32 correct to three decimal places​

Answer 1

Answer:

$\sqrt[4]{32} =2.378$  The 4th root of 32 correct up to 3 decimal places is 2.378.

Step-by-step explanation:

$Given\\x_0=2 and x_1=3\\Let \\x^4=32 \\\\then \\x^4-32=0.\\f(x)=x^4-32\\x_0=2 and x_1=3\\f(2)=2^4-32=-16 < 0\\f(3)=3^4-32=49 > 0\\Now, f(x_0 )f(x_1 ) < 0$

So the root lies between 2 and 3. Using the formulae of false position method

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2- f(2)\frac{(3-2)}{f(3)-f(2)}\\\\x_2=2-(-16)\frac{(3-2)}{49-(-16)}\\\\x_2=2.24615\\f(x_2) = f(2.24615)\\ =-6.546 < 0$

now  $f(x_1)f(x_2) < 0$. So the root lies between $x_0=$2.24615 and $x_1$=3.

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2.24615- f(2.24615)\frac{(3-2.24615)}{f(3)-f(2.24615)}\\\\x_2=2-(-6.546)\frac{(3-2)}{49-(-6.546)}\\\\x_2=2.33499\\f(x_2) = f(2.33499)\\ =-2.27375 < 0$

now  $f(x_1)f(x_2) < 0$. So the root lies between $x_0=$2.33499 and $x_1$=3

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2.33419- f(2.33419)\frac{(3-2.33419)}{f(3)-f(2.33419)}\\\\x_2=2-(-2.27375)\frac{(3-2.33419)}{49-(-2.27375)}\\\\x_2=2.33499\\f(x_2) = f(2.36448)\\ =-0.7433 < 0$

now  $f(x_1)f(x_2) < 0$. So the root lies between $x_0=$2.36448 and $x_1$=3

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2.36448- f(2.36448)\frac{(3-2.36448)}{f(3)-f(2.36448)}\\\\x_2=2-(-0.74339)\frac{(3-2.36448)}{49-(-0.74339)}\\\\x_2=2.33499\\f(x_2) = f(2.37397)\\ =-0.23850 < 0$

now  $f(x_1)f(x_2) < 0$. So the root lies between $x_0=$2.37397 and $x_1$=3

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2.37379- f(2.37379)\frac{(3-2.37379)}{f(3)-f(2.37379)}\\\\x_2=2-(-0.23850)\frac{(3-2.37379)}{49-(-0.23850)}\\\\x_2=2.37700\\f(x_2) = f(2.37700)\\ =-0.07604 < 0$

now  $f(x_1)f(x_2) < 0$. So the root lies between $x_0=$2.37700 and $x_1$=3

$x_2=x_0-f(x_0)\frac{(x_1-x_0)}{f(x_1)-f(x_0)}\\\\x_2=2.37700- f(2.37700)\frac{(3-2.37700)}{f(3)-f(2.37700)}\\\\x_2=2-(-0.07604)\frac{(3-2.37700)}{49-(-0.07604)}\\\\x_2=2.37796\\f(x_2) = f(2.37796)\\ =-0.024438 < 0$

Since after the 6th iterations$f(2.378)$≈.024

$\sqrt[4]{32} =2.378$  hence the 4th root of 32 . correct up to 3 decimal places is 2.378.

Hence, the 4 th root of 32 correct up to 3 decimal places is 2.378.