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Sagot :
The approximate solution to the equation [tex]\dfrac{4}{x - 5} = \sqrt{x + 3} +2[/tex] found using
the Newton-Raphson method is option B.
B. 5.81
How can the Newton-Raphson method be used?
The given function is presented as follows;
[tex]\dfrac{4}{x - 5} = \mathbf{ \sqrt{x + 3} +2}[/tex]
Which gives;
[tex]\dfrac{4}{x - 5} -2= \sqrt{x + 3}[/tex]
[tex]\left(\dfrac{14-2\cdot x}{x-5} \right)^2 = \sqrt{x+3} ^2 = x+3[/tex]
[tex]\mathbf{\dfrac{4\cdot x^2 - 56 \cdot x + 196}{x^2-10\cdot x + 25}} = x + 3[/tex]
Therefore;
[tex]\mathbf{\dfrac{4\cdot x^2 - 56 \cdot x + 196}{x^2-10\cdot x + 25} - (x + 3)} = 0[/tex]
[tex]\dfrac{- x^3 + 11 \cdot x^2 - 51 \cdot x + 121}{x^2-10\cdot x + 25} = 0[/tex]
x³ - 11·x² + 51·x - 121 = 0
Using the Newton Raphson method, with x₀ = 6, we have;
[tex]x_1 = x_0- \dfrac{x^3 - 11 \cdot x^2 + 51 \cdot x - 121}{3 \cdot x^2 - 22 \cdot x + 51}[/tex]
Which gives;
[tex]x_1 = 6- \dfrac{6^3 - 11 \times 6^2 + 51 \times 6 - 121}{3 \times 6^2 - 22 \times 6 + 51} \approx \mathbf{ 5.81}[/tex]
The correct option is B. 5.81
Learn more about the roots polynomial functions here:
https://brainly.com/question/25956931
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