To obtain a root correct to three decimal places of the equation X^3+x²+x+7=0 using the method of false position, we first need to find two initial approximations for the root. We can use the interval [0, 1] since the function changes sign within this interval.
Let f(x) = x^3 + x^2 + x + 7.
When x=0, f(0) = 7, which is positive.
When x=1, f(1) = 10, which is positive.
Since the function changes sign between x=0 and x=1, we can proceed with the false position method.
First, calculate the value of x such that f(x)=0 (i.e. find the root):
x = -f(x) / (f(1) - f(0))
x = -7 / (10 - 7)
x = -7/3
Now we check where the root lies by comparing the sign of f(x) at x=0 and x=-7/3:
f(0) = 7, which is positive
f(-7/3) = (-7/3)^3 + (-7/3)^2 - 7/3 + 7 = -11.9259, which is negative
Since the sign changes from positive to negative, the root lies between x=0 and x=-7/3.
Now, we can calculate the approximations for the root by using the false position method:
x(0) = -7/3, x(1) = 0
x = x(0) - ((f(x(0)) * (x(1) - x(0))) / (f(x(1)) - f(x(0)))
x = -7/3 - ((f(-7/3) * (0 - (-7/3))) / (f(0) - f(-7/3)))
x = -7/3 - ((-11.9259 * 7/3) / (7 - (-11.9259)))
x ≈ -7/3 + 1.9318
x ≈ -1.0681
Now, we have narrowed down the root to the approximate value of x = -1.0681.
To obtain the root correct to three decimal places, we repeat the process with the new interval [0, -7/3]:
x(0) = -7/3, x(1) = -1.0681
Repeat the calculation and we get:
x ≈ -1.178
Therefore, the root of the equation X^3+x²+x+7=0 correct to three decimal places is approximately -1.178.
Use the method of false position to obtain a root correct to three decimal places of each of the following equations X^3+x²+x+7=0
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