Math Problem Statement

Find real root of the equation x^2-5x-7=0 which lies between 2and 3 correct to four decimal places

Solution

We are tasked with finding the real root of the equation x25x7=0x^2 - 5x - 7 = 0 that lies between 2 and 3, correct to four decimal places.

We can solve this equation using the Bisection Method, which is a numerical method for finding roots of continuous functions. Since we know that the root lies between 2 and 3, we can apply this method iteratively to approximate the root.

Steps of the Bisection Method:

  1. Initial Interval: Start with the interval [a,b][a, b], where a=2a = 2 and b=3b = 3.

  2. Midpoint: Compute the midpoint mm of the interval:
    m=a+b2m = \frac{a + b}{2}

  3. Function Evaluation: Evaluate the function f(x)=x25x7f(x) = x^2 - 5x - 7 at the midpoint mm.
    If f(m)=0f(m) = 0, then mm is the root.

  4. Interval Update: If f(a)f(a) and f(m)f(m) have opposite signs, the root lies in [a,m][a, m], so we set b=mb = m. If f(m)f(m) and f(b)f(b) have opposite signs, the root lies in [m,b][m, b], so we set a=ma = m.

  5. Repeat the Process: Continue narrowing the interval by updating aa and bb until the difference between aa and bb is sufficiently small (i.e., until the approximation reaches four decimal places).

Let me calculate the root step by step using the Bisection Method.


Iterative Steps:

  1. Interval [2, 3]:
    f(2)=225(2)7=4107=13f(2) = 2^2 - 5(2) - 7 = 4 - 10 - 7 = -13
    f(3)=325(3)7=9157=13f(3) = 3^2 - 5(3) - 7 = 9 - 15 - 7 = -13

Since the two values give same negative signs, move to next

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Math Problem Analysis

Mathematical Concepts

Algebra
Root-Finding
Numerical Methods
Bisection Method

Formulas

x^2 - 5x - 7 = 0

Theorems

Bisection Method for Root-Finding

Suitable Grade Level

Grades 10-12