1) Examining the following quartic equation:
[tex]-15x^4+4x^3+10=8.245[/tex]We can notice that this an incomplete equation. So let's proceed:
[tex]\begin{gathered} -15x^4+4x^3+10-8.245=0 \\ -15x^4+4x^3+1.755=0 \\ \end{gathered}[/tex]2) Let's use the Newton Method, or the Newton-Raphson method to find the derivative of that equation:
Notice that we need to find the derivative of that quartic function:
f'(x)= -60x³+12x² Applying the power rule
Let's take the first root to be 1, so x_0=1 Now we can plug into the formula
[tex]\begin{gathered} x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)} \\ f^{\prime}(x)=-60x^3+12x^2 \\ x_0=1 \\ x_1=1+\frac{-15x^4+4x^3+1.755}{-60x^3+12x^2} \\ x_1=1+\frac{-15(1)+4(1)+1.755}{-60(1)+12(1)} \\ x_1\approx0.8073 \\ x_2=0.8073+\frac{-15(0.8073)^4+4(0.8073)^3+1.755}{-60(0.8073)^3+12(0.8073)^2} \\ x_2\approx0.10582 \\ x_3=0.10582+\frac{-15(0.10582)^4+4(0.10582)^3+1.755}{-60(0.10582)^3+12(0.10582)^2} \\ x_3\approx0.66797 \\ x_4=0.66797+\frac{-15(0.66797)^4+4(0.66797)^3+1.755}{-60(0.66797)^3+12(0.66797)^2} \\ x\approx0.066485 \end{gathered}[/tex]3) Visualizing graphically
Hence, the answers are:
[tex]x_1=0.66482,x_2\approx-0.52803[/tex]
