The quadratic equation x2-5x+5=0 is to be solved numerically starting with the initial guess x0 = 2. The Newton-Raphson method is applied once to get a new estimate and then the secant method is applied once using the initial guess and this new estimate. The estimated value of the root after the application of the secant method is (round up to 1 decimal place)Correct answer is between '1.4,1.6'. Can you explain this answer?

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Solving the Quadratic Equation

To solve the quadratic equation x^2 - 5x + 5 = 0 numerically, we will use the Newton-Raphson method and then the secant method. The initial guess for both methods is x0 = 2.

Newton-Raphson Method

The Newton-Raphson method is an iterative numerical method used to find the roots of a function. It is based on the idea of approximating a function by its tangent line at a given point and finding the x-intercept of that tangent line.

To apply the Newton-Raphson method, we need to calculate the derivative of the function. In this case, the derivative of f(x) = x^2 - 5x + 5 is f'(x) = 2x - 5.

The formula for the Newton-Raphson iteration is:
x1 = x0 - f(x0) / f'(x0)

Using the initial guess x0 = 2, we can calculate the first iteration as follows:
x1 = 2 - (2^2 - 5*2 + 5) / (2*2 - 5) = 2 - (4 - 10 + 5) / (4 - 5) = 2 - (-1) / -1 = 2 + 1 = 3

Therefore, the new estimate after applying the Newton-Raphson method once is x1 = 3.

Secant Method

The secant method is another iterative numerical method used to find the roots of a function. It is based on the idea of approximating a function by a secant line through two given points and finding the x-intercept of that secant line.

The formula for the secant iteration is:
x2 = x1 - f(x1) * (x1 - x0) / (f(x1) - f(x0))

Using the initial guess x0 = 2 and the new estimate x1 = 3, we can calculate the secant iteration as follows:
x2 = 3 - (3^2 - 5*3 + 5) * (3 - 2) / ((3^2 - 5*3 + 5) - (2^2 - 5*2 + 5)) = 3 - (9 - 15 + 5) * (3 - 2) / ((9 - 15 + 5) - (4 - 10 + 5)) = 3 - (-1) * 1 / (-1 - (-1)) = 3 - (-1) / 0 = 3 - undefined

Since the denominator of the secant iteration is 0, the secant method cannot be applied further.

Conclusion

After applying the secant method once using the initial guess x0 = 2 and the new estimate x1 = 3, we cannot obtain a new estimate due to a division by zero. Therefore, the estimated value of the root after the application of the secant method is undefined.

The correct answer is between 1.4 and 1.6, which indicates that the root lies between these two values. However, since the secant method could not provide a new estimate, we cannot determine the exact value of the root using this method.

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