Solving Quadratic Equations - Algebra 2

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

This problem could be worked out using the quadratic formula, but in this particular case it's easier to factor the left side.

and are the roots that zero the expression on the left side of the equation. In the graph, the curve - which happens to be a paraboloa - will cross the x-axis at the roots.

A few more points...

Observe that the coefficient for the term in the original quadratic is the sum of and . Also, the constant term in the originl equation is the product of and . It's a good rule of thumb to look for numbers that will satsify these conditions when you are setting off to solve a quadratic. Observe how this happens,

If you notice this pattern in a quadratic, then factoring is always a faster approach. The quadratic formula will always work too, but may take a little longer.

Unfortunately you will often find that factoring is not an option since you will not always be abe to easily find such a pattern for most quadratics, especially if the roots are not whole number integers, or if one or both of the roots are complex numbers.

The roots are the values of for which:

                                                                                                                               Reminder

Recall that for a quadratic the general formula for the solution in terms of the constant coefficients is given by:

                                                                                                                             Use the quadratic formula to find the roots.

Notice that is not a real number, and therefore the roots will be complex numbers.

Using the definition of the imaginary unit we can rewrite as follows,

Now we can write the solutions to this problem in the form:

In order to complete the square, we must get all the terms with x in them to one side of the equation. For this problem, we subtract 9 on both sides in order to yield:

.

Remember that completing the square means that we take half of the coefficient of x (), square this new value (), and add it to both sides of the equation. Our equation now looks like:

The left side is a perfect square polynomial, as we have set it up this way. We factor it as such:

After some cleanup, we arrive at:

In order to solve for x, we must take the square root of both sides of the equation.

Finally, we add 6 to both sides of the equation, and simplify the square root of 27:

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

This problem could be worked out using the quadratic formula, but in this particular case it's easier to factor the left side.

and are the roots that zero the expression on the left side of the equation. In the graph, the curve - which happens to be a paraboloa - will cross the x-axis at the roots.

A few more points...

Observe that the coefficient for the term in the original quadratic is the sum of and . Also, the constant term in the originl equation is the product of and . It's a good rule of thumb to look for numbers that will satsify these conditions when you are setting off to solve a quadratic. Observe how this happens,

If you notice this pattern in a quadratic, then factoring is always a faster approach. The quadratic formula will always work too, but may take a little longer.

Unfortunately you will often find that factoring is not an option since you will not always be abe to easily find such a pattern for most quadratics, especially if the roots are not whole number integers, or if one or both of the roots are complex numbers.

The roots are the values of for which:

                                                                                                                               Reminder

Recall that for a quadratic the general formula for the solution in terms of the constant coefficients is given by:

                                                                                                                             Use the quadratic formula to find the roots.

Notice that is not a real number, and therefore the roots will be complex numbers.

Using the definition of the imaginary unit we can rewrite as follows,

Now we can write the solutions to this problem in the form:

Starting with the original equation, we move the constant term of the quadratic over to the other side, so we can set up our "completing the square."

Since the x^2 coefficient is already 1, we don't have to do any division. We can go straight to completing the square by dividing the x coefficient by 2, squaring the result, and adding that result to both sides.

Since the left side is now a perfect square, we can rewrite it as a squared binomial.

Now take the square root of both sides and solve for x.

To solve the quadratic equation, , we set the equation equal to zero and then factor the quadratic, . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations and to obtain the answers and .

This problem could be worked out using the quadratic formula, but in this particular case it's easier to factor the left side.

and are the roots that zero the expression on the left side of the equation. In the graph, the curve - which happens to be a paraboloa - will cross the x-axis at the roots.

A few more points...

Observe that the coefficient for the term in the original quadratic is the sum of and . Also, the constant term in the originl equation is the product of and . It's a good rule of thumb to look for numbers that will satsify these conditions when you are setting off to solve a quadratic. Observe how this happens,

If you notice this pattern in a quadratic, then factoring is always a faster approach. The quadratic formula will always work too, but may take a little longer.

Unfortunately you will often find that factoring is not an option since you will not always be abe to easily find such a pattern for most quadratics, especially if the roots are not whole number integers, or if one or both of the roots are complex numbers.

First we want to put the equation into standard form; we do this by making sure the equation is = 0, so let's subtract 4 from both sides.

We could go straight to the quadratic formula from here, but quadratics are always easier to solve if we eliminate Greatest Common Factors first. In this case the GCF is 4 so let's divide both sides by 4.

Now we can compare against the standard form to find a, b, and c.

By pattern matching, we see that a = 1, b = -3, and c = 4. Now we substitute into the quadratic formula.

We can evaluate the square root of a negative number by factoring out the square root of -1 and calling it the imaginary number i. This gives our answer.

Starting with the original equation, we move the constant term of the quadratic over to the other side, so we can set up our "completing the square."

Since the x^2 coefficient is already 1, we don't have to do any division. We can go straight to completing the square by dividing the x coefficient by 2, squaring the result, and adding that result to both sides.

Since the left side is now a perfect square, we can rewrite it as a squared binomial.

Now take the square root of both sides and solve for x.

To solve the quadratic equation, , we set the equation equal to zero and then factor the quadratic, . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations and to obtain the answers and .

This problem could be worked out using the quadratic formula, but in this particular case it's easier to factor the left side.

and are the roots that zero the expression on the left side of the equation. In the graph, the curve - which happens to be a paraboloa - will cross the x-axis at the roots.

A few more points...

Observe that the coefficient for the term in the original quadratic is the sum of and . Also, the constant term in the originl equation is the product of and . It's a good rule of thumb to look for numbers that will satsify these conditions when you are setting off to solve a quadratic. Observe how this happens,

If you notice this pattern in a quadratic, then factoring is always a faster approach. The quadratic formula will always work too, but may take a little longer.

Unfortunately you will often find that factoring is not an option since you will not always be abe to easily find such a pattern for most quadratics, especially if the roots are not whole number integers, or if one or both of the roots are complex numbers.

First we want to put the equation into standard form; we do this by making sure the equation is = 0, so let's subtract 4 from both sides.

We could go straight to the quadratic formula from here, but quadratics are always easier to solve if we eliminate Greatest Common Factors first. In this case the GCF is 4 so let's divide both sides by 4.

Now we can compare against the standard form to find a, b, and c.

By pattern matching, we see that a = 1, b = -3, and c = 4. Now we substitute into the quadratic formula.

We can evaluate the square root of a negative number by factoring out the square root of -1 and calling it the imaginary number i. This gives our answer.

We begin by designating values of a, b and c by comparing the equation to the standard form.

By pattern matching it is clear that a = 2, b = -3, and c = -5. We can now substitute into the quadratic formula:

Take note that in the two cases where negatives/minus signs are multiplied together \[-(-3) and -4(2)(-5)\], they bceome positive:

Now we simplify and evaluate.

Note that in the previous step we listed the subtraction instance first, as that instance yields the smaller number and it is usually convenient to start sets with smaller numbers first.

To solve the quadratic equation, , we set the equation equal to zero and then factor the quadratic, . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations and to obtain the answers and .

First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.

Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.

So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.

We identify common factors to "pull" out of each group.

And now we factor out x-3.

Setting each factor equal to 0 lets us solve for x.

So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.

First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.

Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.

So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.

We identify common factors to "pull" out of each group.

And now we factor out x-3.

Setting each factor equal to 0 lets us solve for x.

So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.