Question

Solve the equation without using a calculator


`\boldsymbol{4x√(2x-x^2) =2x-1}`

Answer 1

x = 1.92888 (5 d.p.)

Explanation:

Given equation:

`4x√(2x-x^2)=2x-1`

Square both sides:

`\implies (4x√(2x-x^2))^2=(2x-1)^2`

Simplify:

`\implies16x^2(2x-x^2)=4x^2-4x+1`

`\implies32x^3-16x^4=4x^2-4x+1`

`\implies -16x^4+32x^3-4x^2+4x-1=0`

Solve using the Newton-Rhapson method.

`\boxed{\begin{minipage}{8 cm}\underline{The Newton-Rhapson iteration for solving f$(x) = 0$}\\\\$x_(n+1)=x_n-\frac{\text{f}\left(x_n\right)}{\text{f}\:'\left(x_n\right)}$\\\end{minipage}}`

If f(x) = 0 then:

`\text{f}(x)= -16x^4+32x^3-4x^2+4x-1`

Differentiate f(x) to find f'(x):

`\implies \text{f}\:'(x)=-64x^3+96x^2-8x+4`

This means the iteration formula is:

`x_(n+1)=x_n-\frac{-16{x_n}^4+32{x_n}^3-4{x_n}^2+4{x_n}-1}{-64{x_n}^3+96{x_n}^2-8{x_n}+4}`

Let x₀ = 2.

Substitute this into the formula to find x₁:

`\begin{aligned} \implies x_1&=2-(-16(2)^4+32(2)^3-4(2)^2+4(2)-1)/(-64(2)^3+96(2)^2-8(2)+4)\\ & =2-(-9)/(-140)\\&=1.935714286\end{aligned}`

Substitute x₁ into the iteration formula to find x₂:

`\implies x_2=1.928947473`

Repeat until the solution is found:

`\implies x_3=1.928876009`

`\implies x_4=1.928876002`

`\implies x_5=1.928876002`

Therefore, the solution to the given equation is x = 1.92888 (5 d.p.).