For the following problems, use the zero-factor property to solve the equations. (y+6)(2 y+1)=0

Name: Problem 49, Chapter 10: Elementary Algebra
Uploaded: 2022-05-30T15:52:44-08:00
Duration: 3 min 48 s
Channel: Laurie Buchanan
Description: For the following problems, use the zero-factor property to solve the equations. (y+6)(2 y+1)=0

For the following problems, use the zero-factor property to...

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Key Concepts

Zero Product Property

The zero product property states that if the product of two or more factors is zero, then at least one of the factors must be zero. This principle is fundamental when solving polynomial equations because it allows one to set each factor equal to zero, leading directly to the solutions of the equation.

Factoring

Factoring involves rewriting a polynomial or an expression as a product of simpler expressions. It is a valuable technique in algebra that simplifies the process of solving equations by breaking complex expressions into manageable parts, ultimately allowing for the application of the zero product property.

Solving Linear Equations

Solving linear equations is the process of finding the value of the variable that satisfies an equation of the first degree. In the context of applying the zero product property, each individual factor set to zero typically results in a simple linear equation that can be solved using basic algebraic techniques.

Transcript

00:01 All right, here we've got one -third y -squared minus y -sixth equal zero.

00:13 And we want to solve this equation using the quadratic formula.

00:18 So even though we can use fractions in the quadratic formula, i think it's a little bit easier if our coefficients are all integers.

00:27 So i'm going to multiply both sides of this equation by six.

00:32 And that's going to give us 6 times 1 3rd y squared is going to be 2 y squared.

00:45 And 6 times y is 6 y.

00:49 So we're going to have minus 6 y.

00:52 And then 6 times that minus 1 6 is going to give us minus 1.

01:00 And that's just equal to 0 times 6, which is 0.

01:03 But now we've got a little bit nicer numbers to work with here.

01:07 So i'm going to next identify the values of a, b, and c for our quadratic formula.

01:13 So a is the coefficient of our y -squared term.

01:16 A is 2.

01:19 B is negative 6.

01:23 And c is negative 1.

01:26 And now we can substitute those values into the quadratic formula to solve for y.

01:32 So y equals negative b, that's going to be negative, negative six, plus or minus b squared, negative six squared, minus four times a, which is two, times c, which is negative one...

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