Use factoring to solve the equation for real values of the variable. (Enter your answers as a comma-separated list.)
[tex]2 y^{-2}+9 y^{-1}-5=0[/tex]
Answer (2)
Substitute x = y − 1 to transform the equation into a quadratic equation: 2 x 2 + 9 x − 5 = 0 .
Factor the quadratic equation: ( 2 x − 1 ) ( x + 5 ) = 0 .
Solve for x : x = 2 1 or x = − 5 .
Substitute back to find y : y = 2 or y = − 5 1 . The final answer is 2 , − 5 1 .
Explanation
Understanding the Problem We are given the equation 2 y − 2 + 9 y − 1 − 5 = 0 and asked to solve for the real values of y using factoring. This equation involves negative exponents, which can be a bit tricky to work with directly. To simplify things, we'll use a substitution to transform the equation into a more familiar quadratic form.
Making a Substitution Let's make a substitution: let x = y − 1 . This means x = y 1 . Now we can rewrite the original equation in terms of x : 2 x 2 + 9 x − 5 = 0 This is a quadratic equation that we can solve by factoring.
Factoring the Quadratic Equation Now, let's factor the quadratic equation 2 x 2 + 9 x − 5 = 0 . We are looking for two numbers that multiply to 2 × − 5 = − 10 and add up to 9 . These numbers are 10 and − 1 . So we can rewrite the middle term as 10 x − x : 2 x 2 + 10 x − x − 5 = 0 Now, we can factor by grouping: 2 x ( x + 5 ) − 1 ( x + 5 ) = 0 (2x - 1)(x + 5) = 0
Solving for x Now we solve for x by setting each factor equal to zero:
2 x − 1 = 0 ⟹ 2 x = 1 ⟹ x = 2 1
x + 5 = 0 ⟹ x = − 5
So, we have two possible values for x : x = 2 1 and x = − 5 .
Substituting Back and Solving for y Now we need to substitute y − 1 back in for x and solve for y . Recall that x = y 1 .
If x = 2 1 , then y 1 = 2 1 ⟹ y = 2 .
If x = − 5 , then y 1 = − 5 ⟹ y = − 5 1 .
So, the two possible values for y are 2 and − 5 1 .
Final Answer Therefore, the solutions to the equation 2 y − 2 + 9 y − 1 − 5 = 0 are y = 2 and y = − 5 1 .
Examples
Factoring quadratic equations is a fundamental skill in algebra with many real-world applications. For instance, engineers use quadratic equations to model the trajectory of projectiles, such as designing the path of a rocket or the flight of a ball. Similarly, economists use quadratic equations to model cost and revenue curves to optimize business decisions. By understanding how to factor and solve these equations, professionals can make informed predictions and solve complex problems in their respective fields.
The equation 2 y − 2 + 9 y − 1 − 5 = 0 can be transformed using substitution to 2 x 2 + 9 x − 5 = 0 . After factoring, we solve for x and substitute back to find the solutions for y, which are 2 and − 5 1 . The final answers are 2 , − 5 1 .
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