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Methods for Solving Quadratic Equations

Here are the methods we have learned so far to solve quadratic equations. We first started by reviewing factoring. Then we looked at the root method, and finally, we completed the square. Factoring is perhaps the easiest method if we can factor a quadratic equation. If we can't factor it, then we'll want to use one of these two approaches in order to get our solution. And if none of these three methods will work, we have one last method that will always work. That method is the quadratic formula. There are some advantages and disadvantages to this method. Let's see what they are. For any equation written in this form, we can find the solutions to it by doing negative b plus or minus the square root of b squared minus 4ac, all divided by 2a. This is our quadratic formula and it only requires that we know the values of a, b, and c. a is the coefficient of the x squared term, b is the coefficient of the x term, and c would be the constant term. I mentioned the pros and cons of this quadratic formula. Well, the pro is that it can solve any quadratic equation, so long as we have the a, b, and c, and the left-hand side is set equal to 0. We could also have the equation reversed, the left-hand side could be 0 and the right-hand side could be this. For the cons, it's actually very easy to make a sign error in two places. Here, we need to take the negative value of b, so if to change the sign. The other place that we might make a sign error is underneath the radical. b squared should always end up being a positive number, and then we have minus 4ac. Now, depending on the values of a and c, this second term might end up being positive or it might stay negative. You'll want to pay close attention to how many negative signs you have here to determine this number. And finally, another drawback to the quadratic formula is that we have to simplify. We'll need to simplify this radical. And then we'll also need to simplify this fraction if we can. Sometimes, this result will end up being the integers, and sometimes, it will end up being fractions. Keep in mind that we could get imaginary solutions if we have a negative radical here.

The Quadratic Formula Practice 1

When we use the quadratic formula, we want to be sure to use parenthesis. We want to use parenthesis around the values of b, a, and c. For this equation, the value of a is 3, the value of b is negative 4, and the value of c is negative 5. So these are the placeholders for the numbers a, b, and c. You should always do this when you start to solve a quadratic equation using the quadratic formula. This will prevent us from making any sign mistakes with negative b and the minus 4ac. So we plug in the value of b here and here, the value of a here and here, and the value of c here. Now that we've plugged in, it's time to simplify. You try simplifying this expression, and when you think you have your final answer, enter it here. You can either write your answer as two fractions with plus or minus in between them or you can enter in your answer as one giant fraction. The plus or minus symbol will be in the numerator.

The Quadratic Formula Practice 1

The answer is 2 plus or minus the square root of 19, all divided by 3, or we could list this as 2 3rds plus or minus root 19 divided by 3. Either way we write it, these two solutions are correct. We get these answers by first making the negative of b, so this will become positive 4. Negative 4 squared is 16, and these numbers multiplied together are positive 60. We have negative 4 times positive product. This is where I mentioned a sign error usually occurs. You want to watch out to see if you have an even number of negative signs or if you have an odd number of negative signs. We add 16 and 60 to get the square root of 76, and then, we simplify the square root by breaking it down to 4 times 19. The square root of 4 equals 2, and then, we can split up 6 into 2 times 3. This allows us to root a common factor 2 from each of the terms. We know 2 divides into 4 two times, and 2 divides into 2 one time. This leaves us with 2 plus or minus 1 root 19 in our numerator and positive 3 in our denominator.

The Quadratic Formula Practice 2

Try solving this quadratic equation using our quadratic formula. Write your final answer in this box and be sure to use your plus or minus sign. Good luck.

The Quadratic Formula Practice 2

Here, the final solutions are 3 plus or minus 3 times the square root of 2. Great work if you found those answers. The coefficient in front of this x squared term is a 1, since it's not listed. So, a equals 1, b equals negative quadratic formula, being sure to use parenthesis around each of the values. So, we'll have positive 6 for the negative of b, we'll have 36 for squaring b, and then we'll have positive 36 for negative 4acC. Here, the denominator will just be 2, 2 times 1. We have 36 plus 36 in our radical, so really, that's 2 times The 2 stays inside of our radical, and then we simplify it by dividing each of these terms, 2 into 6 equals 3 and 2 into 6 radical 2 is 3 radical 2. Another way to think about this is we could factor 2 from these two 6s, leaving us with this numerator. We cancel the common factors that reduce to 1, leaving us with

The Quadratic Formula Practice 3

Try solving this quadratic equation for the values of r that make it true. Remember, you'll need to set this equation equal to 0 before you can use this quadratic formula. Notice too, that instead of having the variable x, we have the variable r. That's okay though, because it isn't the variable that makes something quadratic. It's the square on the variable. So try rearranging this equation, and then, write your final solutions here. Good luck.

The Quadratic Formula Practice 3

Here, the solutions are 2 plus or minus root 3, all divided by 2. Fantastic work if you figured this one out. We begin to solve by moving this 8r term and this negative 1 to the left-hand side of the equation. So we subtract 8r from both sides, and then, we add positive 1 to both sides as well. Now that we have our quadratic equation set equal to 0, we can use the values of a, b, and c inside of our quadratic equation. a is equal to 4, b is equal to negative 8, and c is equal to positive 1. We substitute these values using parenthesis around each one. The negative of negative 8 is positive 8, negative 8 squared is positive 64, and these three factors give us a product of negative 16. Here, the denominator just simplifies to 8. We subtract inside our radical to get 8 plus or minus the square root of 48, all divided by 8. Next, we want to simplify this radical by splitting it up into 16 times 3. The square root of 16 equals 4, and then we can factor a 4 from our numerator to get 4 times 2 plus or minus 1 root 3. Since a factor of 4 appears in the numerator and in the denominator, we simplify one last time to get 2 plus or minus 1 root 3, all divided by 2. You might have also split this up into two separate fractions and had 2 divided by 2 plus or minus root 3 divided by 2. Of course, this 2 divided by 2 simply reduces to 1. So these would have also been acceptable answers, 1 plus or minus root 3 divided by 2.

The Quadratic Formula Practice 4

Try finding the solutions to this fourth quadratic equation. This time our variable is z. What do you think z could equal?

The Quadratic Formula Practice 4

The solutions for this equation were 2 plus or minus i. Great work if you found these. We start by identifying the variables or the coefficients, a, b, and c. a equals positive 1, b equals negative 4, and c equals positive 5. We plug these into our quadratic formula using parentheses around each of the variables. Here, we'll have the negative of negative 4, which is positive 4. Here, we'll have negative 4 squared, which is 16. And these three factors will give us a product of negative 20. Our denominator will be 2 from this product. And then, we simplify our radical to get the square root of negative 4. The square root of negative 4 is 2i, and then we can factor a 2 from our numerator. So we can simplify one last time. 2 divided by 2 equals 1, and we're left with

The Quadratic Formula Practice 5

Try using the quadratic formula on this equation. What do you think x would be here?

The Quadratic Formula Practice 5

Here, the correct answers were 1 plus or minus root 3, all divided by 2. You may have listed your answer as 1 half plus or minus root 3 divided by 2. These would be correct as well. To solve this equation, we need to first start by setting it equal to 0. So we'll subtract 1 from both sides. We identify a, b, and c using the coefficients in front of each term. a is positive 2, b is negative 2, and c is negative 1. We do the negative of b, which would be positive 2, and then, plus or minus b squared minus 4ac, all divided by 2a. We simplify our answer by cleaning up this radical to get 2 plus or minus root 4 plus 8, all divided by 4. Negative 2 squared is 4 here. And negative 4 times 2 times negative 1 is positive 8. This radical simplifies to root 12, which is 2 times root 3. These two terms share a common factor of 2, so we can rewrite our numerator as 2 times 1 plus or minus 1 root 3. We notice that the numerator and denominator share a common factor of 2, so we simplify that to 1, leaving us with this expression.

Solving Quadratics Choose Your Method 1

Now that we've seen a lot of different types of quadratic equations and how to solve them. I want you to choose the best method in each of the following problems. So, try solving this quadratic equation either by factoring, completing the square, using the root method or the quadratic formula which we just saw. Now here, completing the square would not be the best option. Our a value is 3 and the b value is odd. For completing the square, we want an a value of 1 and a b value that's even. So try one of the other methods and see what you get. For this one, you can enter your answer separated by a comma or you can use the plus or minus symbol. Good luck.

Solving Quadratics Choose Your Method 1

For this quadratic equation, it's best to factor. We know we can factor since there are two factors that multiply to negative 12 and sum to negative 1. Those factors are negative 4 and positive 3. So we can use these as coefficients to rewrite the middle term, and then use factoring by grouping to continue to solve. We set each factor equal to 0, so x could equal negative 1 or x could equal positive 4 3rds. Amazing work if you found these two solutions.

Solving Quadratics Choose Your Method 2

Try tackling this equation. When you think you have the answers, enter them here, separated by a comma, or use the plus or minus sign. Good luck.

Solving Quadratics Choose Your Method 2

The solutions for this equation are x equals 3 plus or minus root 7 all divide by 2. Fantastic work if you found these solutions. I know they were tough. You might have tried factoring here, but it turns out that we can't. For factoring, we would have had to find factors of positive 2 that summed to negative 6. But we know the only factor pairs are positive 2 and positive 1, and negative 2 and negative 1. Now that are those pairs summed to negative 6, so we can't use factoring. We also can't use the root method here. We don't have a perfect square on one side of the equation. Completing the square would also not work, since our a value is positive 2. You would want the a value to be positive 1 and the b value to be even in order to complete the square. So, using this quadratic formula, we have a equal to 2, b equal to negative 6, and c equal to positive 1. We plug the values of a, b, and c into their positions to get this expression. The root of 36 minus 8 is the root of 28, and we know we can simplify this root into 2 root 7. Next, we factor a 2 from the numerator, leaving us with this expression. We cancel the common factor of 2, that appears in the numerator and in the denominator, which leaves us with our most simplified form, our answers. We factor a 2 from these two terms to get this expression. Then, we cancel the common factor of 2 and then the numerator and the denominator to get our final answers.

Solving Quadratics Choose Your Method 3

Here's your third problem. Try finding the solutions to this one and write your answers here.

Solving Quadratics Choose Your Method 3

Here, the solutions are 5 plus or minus root 6, all divided by 2. Amazing work if you got these answers. Here, we want to use the square root method since we have a perfect square on one side of our equation. We square root both sides, indicating the plus or minus on the right. So we'll have 2x minus 5 equals plus or minus root 6. We add 5 to both sides to isolate 2x, and then we divide both sides by 2 to get 1x is equal to 5 halves plus or minus root 6 divided by 2. We can leave our answers separated as two fractions or add and subtract them together to leave them like this.

Solving Quadratics Choose Your Method 4

Here's our last problem for choosing your own method. What do you think would be the best way to solve this equation? Maybe try even solving it two different ways to see if you come up with the same solutions. And what are the solutions? Write them here.

Solving Quadratics Choose Your Method 4

Here, I think it's easiest to complete the square. And it turns out when we solve to complete the square, we get the answers, negative 5 plus or minus root also used the quadratic formula to solve this equation. But since the a here is one and RB is even, I think it's easiest to complete the square. We subtract 3 from both sides, in order to get x squared plus 10x alone on the left-hand side of the equation. Now, we can complete our square. We add b over 2 squared to both sides, which is 25. Now that we've added 25 to both sides, we can rewrite the left side as the quantity x plus 5 squared. The right hand side just simplifies to 22. We take the square root of both sides of the equation to get x plus 5 equals positive or negative root 22. Then, to isolate x, we simply subtract 5 to get our final answers.