Solving Equations Using Factoring | College Algebra

Use factoring to solve. 2. We need to solve given equation by factoring. Let us factor the quadratic. Factoring out GCF of each group.Click here to see ALL problems on Quadratic Equations. Question 77259: the solution set of the equation x^2-4x=0 is? You can put this solution on YOUR website! x^2-4x=0 [factor whereever possible] x(x-4)=0 [set each factor equal to zero and solve for x] x=0 and x-4=0 x=4 . check by...Find solutions for your homework or get textbooks. Problem 9SC from Chapter 3: Use factoring to solve the equations.x2 − 5x + 4 = 0. Get solutions.To solve quadratics by factoring, we use something called "the Zero-Product Property". This property says something that seems fairly obvious, but only after it's been In particular, we can set each of the factors equal to zero, and solve the resulting equation for one solution of the original equation.This article will learn how to solve linear equations using the commonly used methods, namely substitution and elimination. Substitution is a method of solving linear equations in which a variable in one equation is isolated and Solution. Make x the subject of the formula in the second equation.

SOLUTION: the solution set of the equation x^2-4x=0 is?

The given equation reduces to : Factoring the equation we have: (y-7)(y+2)=0. Solving the equations for y x4-5x2-14=0.An old video where Sal solves a bunch of quadratic equations by using factorization methods.Using the quadratic equation x2 - 6x + 8 = 0, perform the following tasks:a) Solve by factoring. The integers that were multiplied together are called the factors of the product. For example In other words, 4 and 2 are the ROOTS or SOLUTION to your ORIGINAL question. Is this clear?Use the system of equations to augment the coefficient matrix and the constant matrix. Systems of linear equations can be solved by first putting the augmented matrix for the system in reduced To find the solutions (if any) to the original system of equations, convert the reduced row-echelon...

Solved: Use factoring to solve the equations.x2 5x + 4 = 0

3 OBJECTIVE Solve quadratic equations by factoring. 4 ZERO - PRODUCT PROPERTY For any real numbers a and b, if ab = 0, then a = 0 or b = 0. Example: If (x + 3)(x + 2) = 0, then x + 3 = 0 or x + 2 = 0. 5 EXAMPLE 1 - USING THE ZERO-PRODUCT PROPERTY What are the solutions of the...Factoring Quadratic Equations when a = 1. A quadratic equation is an equation that contains a Now that the equation has been factored, solve for x. Step 4: Set each factor to zero and solve In these cases it is usually better to solve by completing the square or using the quadratic formula.Use factoring to solve. Answers (1). Use factoring to solve" in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.Solve the following system of equations: x+y=7, x+2y=11. Here are more examples of how to solve systems of equations in Algebra Calculator. Feel free to try them now.Solve x^2 - 5x - 14 = 0. • 335 тыс. просмотров 2 года назад. Solving Quadratic Equations using the Quadratic Formula - Example 3. patrickJMT.

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This lesson covers some ways to remedy quadratics, similar to taking square roots, completing the sq., and the use of the Quadratic Formula. But we'll get started with solving through factoring.

(Before achieving the subject of solving quadratic equations, you must already know the way to element quadratic expressions. If not, first review how to component quadratics.)

You've already factored quadratic expressions. The new thing here is that the quadratic expression is part of an equation, and you might be told to clear up for the values of the variable that make the equation true. Here's how it works:

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Solve (x – 3)(x – 4) = 0 by factoring.

Okay, this quadratic is already factored for me. But how do I use this factorisation to remedy the equation?

To solve quadratics via factoring, we use something known as "the Zero-Product Property". This assets says one thing that turns out somewhat obvious, but most effective after it is been pointed out to us; namely:

Zero-Product Property: If we multiply two (or extra) issues together and the result's equal to zero, then we know that at least one of the ones things that we multiplied should even have been equal to zero. Put differently, the most effective approach for us to get 0 once we multiply two (or extra) components in combination is for one of the elements to had been 0.

So, if we multiply two (or extra) components and get a zero outcome, then we all know that no less than one of the components used to be itself equivalent to zero. In explicit, we can set each and every of the elements equivalent to zero, and remedy the resulting equation for one answer of the authentic equation.

We can handiest draw the helpful conclusion about the factors (namely, that one of the ones factors must were equal to zero, so we will set the elements equivalent to zero) if the product itself equals zero. If the product of components is equal to the rest non-zero, then we will be able to not make any claim about the values of the elements.

Therefore, when solving quadratic equations by means of factoring, we should always have the equation in the form "(quadratic expression) equals (zero)" earlier than we make any attempt to solve the quadratic equation by factoring.

Returning to the workout:

The Zero Factor Principle tells me that at least one of the elements will have to be equivalent to zero. Since a minimum of one of the components should be zero, then I can set each of the elements equal to zero:

x – 3 = 0   or   x – 4 = 0

This offers me easy linear equations, and they're simple to solve:

x = 3   or   x = 4

And these two values are the answer they are searching for:

Note that "x = 3, 4" manner the similar thing as "x = 3 or x = 4"; the best distinction is the formatting. The "x = 3, 4" format is extra not unusual.

Solve x2 + 5x + 6 = 0, and test.

This equation is already in the shape "(quadratic) equals (zero)" however, not like the earlier instance, this isn't but factored. I MUST aspect the quadratic first, because it is just once I MULTIPLY and get 0 that I can say the rest about the factors and solutions. I will be able to't conclude anything about the individual phrases of the unfactored quadratic (like the 5x or the 6), because I will be able to upload loads of stuff that totals to zero.

So the very first thing I have to do is component:

x2 + 5x + 6 = (x + 2)(x + 3)

Now I will restate the original equation in phrases of a product of elements, with this product being equivalent to zero:

(x + 2)(x + 3) = 0

Now I can solve every factor by means of environment every one equal to 0 and fixing the resulting linear equations:

x + 2 = 0   or   x + 3 = 0

x = –2   or   x = – 3

These two values are the resolution to the original quadratic equation. So my solution is:

x = –3, –2

I'm not finished, though, as a result of the original exercise informed me to "check", this means that that I need to plug my answers again into the authentic equation, and make sure it comes out correct. In this example, I'll be plugging into the expression on the left-hand facet of the authentic equation, and verifying that I finally end up with the right-hand side; namely, with 0:

checking x = –3:

[–3]2 + 5[–3] + 6

9 – 15 + 6

9 + 6 – 15

15 – 15

0

checking x = –2:

[–2]2 + 5[–2] + 6

4 – 10 + 6

4 + 6 – 10

10 – 10

0

When an workout specifies that you will have to remedy "and check", the above plug-n-chug, they are searching for you to display that you plugged your resolution into the authentic workout and were given something that worked out correct. The above, where I confirmed my tests, is all they're in need of. But do your paintings neatly!

By the means, you can use this "checking" method to check your solutions to any "solving" exercise. So, as an example, if you're not certain of your resolution to a "factor and solve" question on the subsequent take a look at, check out plugging your answers into the authentic equation, and confirming that your solutions lead to true statements.

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Solve x2 – 3 = 2x.

This equation isn't in "(quadratic) equals (zero)" shape, so I will be able to't check out to solve it yet. The very first thing I want to do is get all the phrases over on one aspect, with 0 on the different side. Only then can I ingredient and solve:

x2 – 3 = 2x

x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0, x + 1 = 0

x = 3, x = –1

Then my resolution is:

x = –1, 3

Solve (x + 2)(x + 3) = 12.

It is quite common for students to see this sort of equation, and say:

"Cool! It's already factored! So I'll set the factors equal to 12 and solve to get x = 10 and x = 9. That was easy!"

Yeah, that was once simple; it was once additionally unsuitable. Very, very mistaken.

Besides the incontrovertible fact that neither (10 + 2)(10 + 3) nor (9 + 2)(9 + 3) equals 12, we will have to never forget that we must have "(quadratic) equals (zero)" before we can clear up by means of factoring.

Returning to the workout:

Tempting despite the fact that it may be, I cannot set every of the components on the left-hand aspect of the equation equivalent to the other facet of the equation and resolve. Doing so would give me an entirely-wrong mess.

Instead, I first have to multiply out and simplify the left-hand facet, then subtract the 12 over to the left-hand side, and re-factor. Only then can I remedy.

(x + 2)(x + 3) = 12

x2 + 5x + 6 = 12

x2 + 5x – 6 = 0

(x + 6)(x – 1) = 0

x + 6 = 0, x – 1 = 0

x = –6, x = 1

Then my solution is:

x = –6, 1

Solve x2 + 5x = 0.

This two-term quadratic is easier to aspect than have been the previous quadratics: I see straight away that I can aspect an x out of each terms, taking the x out front. This offers me:

x (x + 5) = 0

A quite common mistake that scholars make at this level is to "solve" the equation for "x + 5 = 0" via dividing through through the x. But that is an invalid step. Why? Because we will be able to't divide by 0. How does that come into play here?

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Dividing via by way of the component x makes the implicit assumption that x used to be not equal to 0. There is de facto no justification for making that assumption! And making that assumption would purpose us to lose part of our resolution to this equation.

Returning to the workout:

I want to remember the fact that it is ok for a factor to comprise only a variable, with out being added to other terms; particularly, "x" is a superbly valid factor. I want to set each of the factors equal to zero, and then resolve the two resulting linear equations:

x(x + 5) = 0

x = 0, x + 5 = 0

x = 0, x = –5

Then my resolution is:

x = 0, –5

The previous instance had two terms and was easy to factor. There is one different case of two-term quadratics that we will be able to aspect to clear up. It's only a bit extra sophisticated:

Solve x2 – 4 = 0.

This equation is in "(quadratic) equals (zero)" form, so it's ready for me to solve by way of factoring. But how do I element this? By noticing that this can be a difference of squares. I'll follow the difference-of-squares system that I've memorized:

x2 – 4 = 0

(x – 2)(x + 2) = 0

x – 2 = 0, x + 2 = 0

x = 2, x = –2

Then my answer is:

x = –2, 2

Note: The answer above will also be formatted as "x = ± 2". This is pronounced as "x is equal to plus or minus 2".

The final instance above leads us into how to resolve through taking square roots, on the next web page.

You can use the Mathway widget below to observe solving quadratic equations via factoring. Try the entered workout, or kind on your own workout. Then click the button and make a selection "Solve by factoring" to compare your solution to Mathway's. (Or skip ahead to the subsequent web page.)

(Click "Tap to view steps" to be taken immediately to the Mathway website online for a paid improve.)

URL: https://www.purplemath.com/modules/solvquad.htm

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