The solution of the equation given by $2 x^{3} + 5 x + A = 0$ , (where A is a constant) is determined by Newton-Raphson method. First assumption for root of the equation is $x_{0} = 2$ If after one iteration the root obtained is $x_{1} = 3.5$ , then the value of A is____
-69.5

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Solution

The correct option is A -69.5
From Newton Raphson method,
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$3.5 = 2 - \frac{2 x_{0}^{3} + 5 x_{0} + A}{6 x_{0}^{2} + 5}$
$1.5 = - \frac{2 (2)^{3} + 5 (2) + A}{6 (2)^{2} + 5}$
$- 1.5 = \frac{16 + 10 + A}{29}$
$- 43.5 = 26 + A$
$A = - 69.5$

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The solution of the equation given by 2x3+5x+A=0, (where A is a constant) is determined by Newton-Raphson method. First assumption for root of the equation is x0=2 If after one iteration the root obtained is x1=3.5, then the value of A is____-69.5

The correct option is A -69.5From Newton Raphson method, x1=x0−f(x0)f′(x0) 3.5=2−2x30+5x0+A6x20+5 1.5=−2(2)3+5(2)+A6(2)2+5 −1.5=16+10+A29 −43.5=26+A A=−69.5

The solution of the equation given by $2 x^{3} + 5 x + A = 0$ , (where A is a constant) is determined by Newton-Raphson method. First assumption for root of the equation is $x_{0} = 2$ If after one iteration the root obtained is $x_{1} = 3.5$ , then the value of A is____
-69.5

The correct option is A -69.5
From Newton Raphson method,
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$3.5 = 2 - \frac{2 x_{0}^{3} + 5 x_{0} + A}{6 x_{0}^{2} + 5}$
$1.5 = - \frac{2 (2)^{3} + 5 (2) + A}{6 (2)^{2} + 5}$
$- 1.5 = \frac{16 + 10 + A}{29}$
$- 43.5 = 26 + A$
$A = - 69.5$