Finding Roots - Algebra 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.

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.

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.

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.

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}.

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}.

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 .

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

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}.

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

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}.