$x^2-10 x+25=0$
$\\frac{1}{3}=m+\\frac{m-1}{m}$
Answer (2)
The solutions to the equations are: x = 5 for the quadratic equation, and m = 3 − 1 ± 10 for the rational equation.
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Solve the quadratic equation x 2 − 10 x + 25 = 0 by factoring to find x = 5 .
Solve the rational equation 3 1 = m + m m − 1 by multiplying both sides by 3 m and simplifying to get a quadratic equation in m .
Use the quadratic formula to solve for m in the equation 3 m 2 + 2 m − 3 = 0 , resulting in m = 3 − 1 ± 10 .
The solutions are x = 5 and m = 3 − 1 ± 10 , which means x = 5 and m = 3 − 1 ± 10 .
Explanation
Problem Analysis We are given two equations:
x 2 − 10 x + 25 = 0
3 1 = m + m m − 1
Our objective is to solve for x in the first equation and for m in the second equation.
Solving for x Let's solve the first equation for x . The equation is a quadratic equation, and we can solve it by factoring. We are looking for two numbers that multiply to 25 and add up to -10. These numbers are -5 and -5. Therefore, we can rewrite the equation as: ( x − 5 ) ( x − 5 ) = 0 ( x − 5 ) 2 = 0
Finding the value of x Taking the square root of both sides, we get: x − 5 = 0 Adding 5 to both sides, we find: x = 5
Solving for m Now, let's solve the second equation for m :
3 1 = m + m m − 1 To eliminate the fractions, we multiply both sides of the equation by 3 m :
3 m ⋅ 3 1 = 3 m ⋅ ( m + m m − 1 ) m = 3 m 2 + 3 ( m − 1 ) m = 3 m 2 + 3 m − 3
Rewriting the equation Rearranging the equation into a standard quadratic form, we get: 3 m 2 + 3 m − 3 − m = 0 3 m 2 + 2 m − 3 = 0
Applying the quadratic formula Now we use the quadratic formula to solve for m . The quadratic formula is given by: m = 2 a − b ± b 2 − 4 a c In our equation, a = 3 , b = 2 , and c = − 3 . Plugging these values into the quadratic formula, we get: m = 2 ( 3 ) − 2 ± 2 2 − 4 ( 3 ) ( − 3 ) m = 6 − 2 ± 4 + 36 m = 6 − 2 ± 40 m = 6 − 2 ± 2 10 m = 3 − 1 ± 10
Finding the values of m Therefore, the solutions for m are: m = 3 − 1 + 10 ≈ 0.720759 m = 3 − 1 − 10 ≈ − 1.387426
Final Answer In summary, we have found the solutions for x and m :
x = 5
m = 3 − 1 + 10 or m = 3 − 1 − 10
Examples
Quadratic equations are used in various real-life scenarios, such as calculating the trajectory of a ball, determining the dimensions of a garden to maximize area, or modeling the growth of a population. Similarly, solving rational equations is crucial in fields like physics and engineering, where they are used to analyze electrical circuits, fluid dynamics, and other complex systems. Understanding these concepts provides a foundation for solving practical problems in science and technology.
For example, imagine you are designing a rectangular garden with a fixed perimeter. Using a quadratic equation, you can determine the dimensions that will yield the maximum area for your garden, ensuring you get the most out of your space. Or, if you're an engineer working with electrical circuits, solving rational equations helps you calculate the current and voltage in different parts of the circuit, ensuring it functions correctly and efficiently.
Let's say you want to fence a rectangular garden using 20 meters of fencing. If one side is 'x' meters, the other side will be (10-x) meters. The area A of the garden can be expressed as A = x(10-x) = 10x - x^2. To find the maximum area, you can set the derivative of A with respect to x to zero, which leads to a quadratic equation. Solving this equation will give you the value of x that maximizes the garden's area.
