A solution to the equation x^3 + 2x - 25 = 0 lies between 2 and 3. Find this solution correct to 1 decimal place

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The solution to the equation x^3 + 2x - 25 = 0, correct to 1 decimal place, is approximately x = 2.7.

To find the solution to the equation x^3 + 2x - 25 = 0, we can use numerical methods such as the bisection method or Newton's method. Let's use the bisection method to find the solution between 2 and 3.

Evaluate the equation at the lower and upper bounds:

f(2) = (2)^3 + 2(2) - 25 = 8 + 4 - 25 = -13

f(3) = (3)^3 + 2(3) - 25 = 27 + 6 - 25 = 8

Check if the function changes sign between the lower and upper bounds:

Since f(2) is negative and f(3) is positive, we can conclude that there is at least one root between 2 and 3.

Perform iterations of the bisection method:

Divide the interval [2, 3] into two equal halves:

Lower bound: a = 2, Upper bound: b = 3

Midpoint: c = (a + b) / 2 = (2 + 3) / 2 = 2.5

Evaluate the function at the midpoint:

f(c) = (2.5)^3 + 2(2.5) - 25 = 15.625 + 5 - 25 = -4.375

Check the sign of f(c):

Since f(c) is negative, we update the lower bound to c, as the solution lies between the lower bound and c.

Repeat the steps with the new interval [2.5, 3]:

Lower bound: a = 2.5, Upper bound: b = 3

Midpoint: c = (a + b) / 2 = (2.5 + 3) / 2 = 2.75

Evaluate the function at the new midpoint:

f(c) = (2.75)^3 + 2(2.75) - 25 = 20.796875 + 5.5 - 25 = 1.296875

Check the sign of f(c):

Since f(c) is positive, we update the upper bound to c, as the solution lies between c and the upper bound.

Repeat the steps until the desired level of precision is achieved:

Continue dividing the interval and evaluating the function at each midpoint until the interval becomes sufficiently small.

Continue the bisection method iterations:

Repeat the steps until the interval is small enough:

[2.5, 2.75]

[2.625, 2.75]

[2.6875, 2.75]

[2.71875, 2.75]

[2.734375, 2.75]

[2.734375, 2.7421875]

The final interval [2.734375, 2.7421875] contains the solution.

Therefore, the solution to the equation x^3 + 2x - 25 = 0, correct to 1 decimal place, is approximately x = 2.7.

To know more about 1 decimal place:

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ERROR ANALYSIS. Describe and correct the error in solving the system of linear equations.
2x+y=5 Equation 1
3x-2y=4 Equation 2
Step 1:
2x+y=5
y= -2x+5
Step 2:
2x+(-2x+5)=5
2x-2x+5=5
5=5
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