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Let's take a look at solving radical equations.
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A radical equation is an equation where the variable appears under the radical sign.
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Here is an example, the square root of x minus 7 equals 3.
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So x appears under the square root sign.
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To solve a radical equation, you need to raise both sides of the equation to a power
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so that you can eliminate the square root sign.
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In this example, we will raise both sides to the second power to eliminate the square root.
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This will give us the square root of 8 x minus 7, that entire quantity squared, equals 3 squared.
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On the left side when you simplify this you will get 8 x minus 7.
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On the right side 3 squared equals 9.
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And now we solve this the usual way for solving a linear equation.
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I will add 7 to both sides and get 8 x minus 7 plus 7 equals 9 plus 7.
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This gives 8 x equals 16. And then to solve for x we will divide both sides by 8,
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giving us 8 x over 8 equals 16 divided by 8.
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Make that an 8.
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So this will simplify to x equals 2.
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You have to remember when you are solving radical equations that you need to check the answers
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because squaring both sides could introduce extra answers.
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So for the check, go back to the original problem, substitute 2 in for x,
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and check to see if when you simplify the left side that you get 3.
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So for the check you will have the square root of 8 times 2 minus 7, and the question is does it equal 3?
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Simplifying, we will have square root of 16 minus 7. Is it 3?
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So the square root of 9, is it 3?
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And you end up with a true statement 3 equals 3.
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Since we have a true statement on the check, x equals 2 is a valid answer.
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Now we will take a look at another radical equation.
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Let's solve the square root of 10 minus 3 x equals x.
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To eliminate the radical, we will square both sides,
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giving us the square root of 10 minus 3 x squared equals x squared.
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This will eliminate the square root and simplify to 10 minus 3 x equals x squared.
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At this point, I need to get 0 on one side of the equation,
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and so that I do not have a negative coefficient in front of x squared I am going to choose to get 0 on the left side.
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This will give me 0 equals x squared plus 3 x minus 10.
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This problem can be factored. It factors into x plus 5 times x minus 2,
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and when you set each factor equal to 0, you get what looks like two answers.
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x equals negative 5 or x equals positive 2.
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But you must remember, because we squared both sides, that you have to check these answers.
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So we will check the first one, x equals negative 5.
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Plug negative 5 back into the original equation.
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Simplify it, and see if you end up with a true statement.
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This will give us the square root of 10 minus 3 times negative 5.
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We want to know, does this equal negative 5?
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When you simplify the left side, you will have the square root of 10 plus 15. Is it negative 5?
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So, the square root of 25. Is it negative 5?
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As it turns out we end up with 5 equals negative 5.
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This is a false statement that we have on the check.
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This means that negative 5 is not an answer; it is not a valid answer.
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It is an extraneous answer.
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I just put a mark through it right here to indicate that it is not a valid answer.
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For the check on x equals two, plug 2 into the original equation.
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And you have 10 minus 3 times 2, the square root of that quantity. Is it 2?
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When you simplify this, you will get the square root of 4. Is it 2?
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And simplifying we get 2 equals 2, a true statement.
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So x equals 2 is the only answer to this problem.
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Practice solving radical equations and remember to check your solutions.
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Let's take a look at solving quadratic type equations.
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We will do this problem 7 x to the fourth minus 6 x squared minus 1 equals 0.
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Now notice the exponents on these terms.
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The first term has an exponent of 4, the second term has an exponent of 2, and the third term is a constant term.
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The exponents, the two being exactly half of 4, characterizes this as a quadratic type equation.
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A lot of these equations can be solved using either factoring or substitution.
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I will work this one both ways. First by factoring.
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We will just treat it as a regular problem where we factor an expression.
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On the left side I can factor 7 x to the fourth into 7 x squared times x squared.
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Then for the 1, I will have to put 1 times 1.
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Then with a little bit of thinking, you will figure you need a minus sign.
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x squared minus one, and this factor is 7 x squared plus 1.
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Now that it is factored, put each factor equal to zero, and solve for x.
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So 7 x squared plus one equals zero or x squared minus 1 equals 0.
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Let’s work the left side first.
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So I have 7 x squared equals negative 1. And when you divide by 7, x squared is negative 1/7.
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We could continue solving this for x by taking the square root of both sides,
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however you will notice you cannot substitute a real number for x, square it, and get a negative 1/7.
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So this factor, 7 x squared plus 1, does not give us a valid solution to this problem.
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For the other factor, we will add one to both sides to get x squared equals one.
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At this point, we can say that x is either plus or minus one.
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So the solution to this problem is right here is plus and minus one.
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Let’s take this one, put an x through it, just to say that is not a solution.
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So, the answer is plus or minus one.
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Now I would like to do the same problem but by using substitution.
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So for substitution I will recopy 7 x to the fourth minus 6 x squared minus 1 equals 0.
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What I want to do is substitute for x squared, and it may be helpful to rewrite this in this form:
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7 x squared, it is x squared squared to get x to the fourth, minus 6 x squared minus 1 equals 0.
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At this point we will substitute u for x squared.
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So, I am going to write let u equal x squared, giving us 7u squared minus 6u minus 1 equals 0.
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Now we will use factoring for 7 u squared minus 6u minus 1 equals 0.
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This will factor into 7 u plus 1 times u minus 1,
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and then putting each factor equal to 0. 7u plus 1 is 0 and u minus 1 equals 0.
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We will have u equals negative 1/7, when I solve this one for u.
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Now the problem here is that I want to know what x is equal to, not what u is equal to, so I have to go back and resubstitute.
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Since u was equal to x squared, I will get x squared equals negative 1/7.
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Again I cannot plug in or substitute a number for x, a real number for x, and square it and get negative 1/7.
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So this is not a valid solution for this problem.
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For this side u equals 1, but you need to resubstitute.
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u equals x squared, so x squared equals 1, and here you will get the solution plus or minus one.
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Now let's take a look at another example.
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Let's take a look at the next example.
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3 x to the 2/3 minus 2 x to the 1/3 minus 1 equals 0.
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This is another equation that could be classified as quadratic type.
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The first exponent, the first term, has an exponent of 2/3,
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the second term has an exponent of 1/3, and the third time is a constant term.
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The two to one ratio on exponents fits with the 2/3 and the 1/3.
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This one I will solve by factoring.
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To factor 3 x to the 2/3 it is simply this: 3 x to the 1/3 times x to the 1/3.
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And then for the 1, of course that will be 1 times 1 is 1.
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Placing in the plus and the minus sign, I will have x to the 1/3 minus 1
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and this one is 3 x to the 1/3 plus 1 for the factors.
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Then put equal to zero on the right side of this equation.
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Now we will put each factor equal to zero.
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3 x to the 1/3 plus 1 equals 0 or we have x to the 1/3 minus 1 equals 0.
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I will work the one on the left side first.
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Subtract 1 from both sides. 3x to the 1/3 equals negative 1.
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When you divide by 3, you have x to the 1/3 equals negative 1/3.
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To solve for x, I will need to raise both sides to the third power.
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This will give us x equals negative 1/3 to the third power. So, x will be negative 1/27.
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So, the factor 3 x to the 1/3 plus 1 equals zero has given us x equals negative 1/27.
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For the other factor, x to the 1/3 minus 1, we have that one equal to 0.
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Add 1 to both sides, x to the 1/3 equals 1.
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When you raise both sides to the third power, you will have x equals 1 to the third power, which is simply x equals 1.
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So the two solutions to this problem are negative 1/27 and positive 1.
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Now we will take a look at another example.
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Let's take a look at the next example.
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This example is quadratic type.
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I have a quantity raised to the second power within the first term.
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In the second term, I have the same quantity and it is raised to the first power
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although we normally do not write the 1 for the exponent.
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This equation you could solve by multiplying x plus 2 times x plus 2,
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distributing the 8, and distributing negative 23, then finding like terms.
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Or we can use substitution. Substitution is the more efficient way to go, so we are going to work with substitution.
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I am going to say let u equal x plus 2, and then with substitution you have 8 u squared minus 23 u minus 3 equals 0.
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Now factoring 8 u squared minus 23 u minus 3 is much easier
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than what I would have gotten had I multiplied all of this out at the beginning.
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To factor 8 u squared minus 23 u minus 3, I will factor it into 8 u times u and of course the 3 will be 3 times 1.
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I will have to put the 3 in the second set of parentheses so that when I multiply I will get the negative 23 u and the middle.
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This would force the negative sign in front of the 3 and a positive sign in front of the 1.
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At this point, I will set each factor equal to 0.
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8 u plus 1 equals 0 or u minus 3 equals 0.
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Solving for u, we will have that u equals negative 1/8 and here u equals 3.
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The problem with this again is that I was supposed to solve for x and not u.
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So I have to resubstitute the value for u. u is x plus 2.
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So instead of u equals negative 1/8, x plus 2 equals negative 1/8.
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And to solve for x subtract 2 from both sides.
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This will give me x equals negative 2 minus 1/8.
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Getting a common denominator, it will be 8. I will have negative 16/8 minus 1/8.
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So one possible solution for x is negative 17 / 8.
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For this one u equals 3, we will put the same thing over here.
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u equals 3, and then resubstitute x plus 2 in for u.
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Subtract 2 from both sides and get x equals 1.
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So the solutions to this equation will consist of negative 17/8 and x equals 1.
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Practice solving these equations until you have mastered the skill.
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Now we will look at other equations that we could solve by factoring.
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We will do this example, x cubed equals 81 x.
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You may be tempted to divide by x on this problem,
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but remember you cannot divide by zero, and we have no guarantee that x is not zero at this point.
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To solve this equation what you need to do instead is subtract 81x from both sides to get x cubed minus 81x equals 0.
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And now we will factor. Start by factoring out an x.
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This will give you x times x squared minus 81 equals 0.
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You could put each factor equal to 0 at this point, or you can continue factoring.
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x squared minus 81 will factor into the difference of two squares, x plus 9 times x minus 9.
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So I have x times x plus 9 times x minus 9 equals 0.
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Set each factor equal to 0, you will get x equals 0, x plus 9 equals 0, and x minus 9 equals 0.
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Solving each one, we will have x equals negative 9, x equals positive 9, or x equals 0.
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The three solutions are 0, negative 9, and 9.
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And now we will look at another example.
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Let's take a look at another example.
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This equation we will solve by grouping.
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The first two terms have a common factor of x squared, so I will factor out the x squared.
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And the last two terms have a common factor of negative 25. Factor out negative 25.
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Notice with the grouping I now have x plus 3 and x plus 3 with each of these terms, so I will factor out x plus 3.
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And then we will have x squared minus 25 for the second factor.
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I could factor x squared minus 25, or at this point I could put both factors equal to 0.
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I am going to do the second one, I am going to set each factor equal to 0.
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x plus 3 equals 0 and x squared minus 25 equals 0.
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For this one move the 3 over, subtract 3 from both sides, and get x equals negative 3.
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For the second one, x squared minus 25 equals 0, add 25 to both sides, and get x squared equals 25.
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When you square root both sides, you will get x is plus or minus 5.
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So, the three solutions to this problem are x equals negative 3 and x equals plus or minus 5.
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Practice solving these problems until you have mastered the skill.