Solve and check each of the equations. x(x+7)-2=28 |...

Name: Problem 10, Chapter 1: Algebra 2 and Trigonometry
Uploaded: 2019-07-30T09:02:12-08:00
Duration: 4 min 16 s
Channel: M S
Description: Solve and check each of the equations. x(x+7)-2=28

Key Concepts

Solution Verification

Once solutions are obtained, it is important to substitute them back into the original equation to verify their validity. This step ensures that no extraneous solutions have been included and confirms the accuracy of the obtained answers.

Expansion of Expressions

This concept involves using the distributive property to eliminate parentheses, turning products like x(x + 7) into x² + 7x. It is a fundamental algebraic skill used to simplify equations and prepare them for further steps in solving.

Standard Form of a Quadratic Equation

Converting an equation into the standard quadratic form, ax² + bx + c = 0, is crucial. This transformation allows for the application of systematic methods such as factoring or the quadratic formula to find the solutions of the equation.

Solving Quadratic Equations

This involves methods such as factoring, completing the square, or using the quadratic formula to find the values of the variable that satisfy the quadratic equation. Understanding these techniques is a key aspect of algebra in dealing with second-degree equations.

Transcript

00:02 So we want to solve and check this equation.

00:06 So the first step would be to get all the numbers on one side so we can set equal to zero.

00:11 So i'm going to go ahead and distribute this x.

00:15 So we have x squared plus 7x and then bring down that minus 2.

00:21 Now we want to subtract 28 from both sides.

00:26 So we'll have x squared plus 7x and negative 2 minus 28 gives us negative 30.

00:34 So we have zero on that side.

00:39 So the first step to factor this polynomial will be to separate this middle term.

00:43 So we know we need it to multiply we need to multiply our coefficient of 1 and negative 30.

00:50 And now we give us negative 30.

00:52 And we know we need to add to positive 7.

00:55 So we list the factors of 30, 1 in 30, 2 and 15, 3 and 10, and 5 and 6.

01:04 And because this 30 is negative, we know that both terms have one of the terms have to be negative so it's either one or negative 30 but not both because that would make our 30 positive so it's not going to be this one it's not going to be this one but it will be our 3 in 10 we just need to figure out which one will get the negative so if we have negative 3 that'll give us positive 7 so this is our combination so we separate this and split apart our middle terms.

01:41 So we have plus 10x minus 3x, then bring down that minus 30.

01:48 So we group the first two and group the second two.

01:52 And we can pull out an x and we're left with x plus 10.

01:55 And then if we pull out negative 3, we're left with x plus 10.

01:59 And we know we're on the right track because this term equals this term.

02:03 So we're left with x plus 10 times x minus 3 and that equals 0.

02:11 So now we want to check and by checking we just plug back in our solution...