Solving Quadratic Equations - Algebra 2
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Solve the following quadratic inequality, and report your answer in interval form:
Solve the following quadratic inequality, and report your answer in interval form:
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The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal.
Negative $x^2$'s are hard to work with, so we multiply through by -1.
Now we can factor easily.
By the zero product property, each of these factors will be equal to 0.
Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality.
Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as:
The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal.
Negative $x^2$'s are hard to work with, so we multiply through by -1.
Now we can factor easily.
By the zero product property, each of these factors will be equal to 0.
Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality.
Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as:
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Solve this inequality.
Solve this inequality.
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Combine like terms first.
Factor
The zeroes are 3 and 8 so a number line can be divided into 3 sections.
X<3 works, 3<x<8 does not work, and x>8 works
Combine like terms first.
Factor
The zeroes are 3 and 8 so a number line can be divided into 3 sections.
X<3 works, 3<x<8 does not work, and x>8 works
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Solve the following quadratic inequality:
Solve the following quadratic inequality:
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First we want to rewrite the quadratic in standard form:
Now we want to set it = 0 and factor and solve like normal.
Using the zero product property, both factors produce a zero:
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
Ergo, 
.
First we want to rewrite the quadratic in standard form:
Now we want to set it = 0 and factor and solve like normal.
Using the zero product property, both factors produce a zero:
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
Ergo,
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Solve the following quadratic inequality:
Solve the following quadratic inequality:
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1. Rewrite the equation in standard form.
2. Set the equation equal to 
and solve by factoring.
So, 
and 
are our zeroes.
3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing 
by plugging it into your original inequality.)
Because the above statement is true, the solution is the interval between 
and 
.
1. Rewrite the equation in standard form.
2. Set the equation equal to
So,
3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing
Because the above statement is true, the solution is the interval between
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What is the discriminant of the following quadratic equation: 
What is the discriminant of the following quadratic equation:
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The discriminant of a quadratic equation in 
form is equal to 
. The given equation is not in that form however, so we must first multiply it out to get it into that form. We therefore obtain:
We therefore have 
, 
, and 
. Our discriminant is therefore:
The correct answer is therefore 
The discriminant of a quadratic equation in
We therefore have
The correct answer is therefore
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Solve:
Solve:
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Start by setting the inequality to zero and by solving for 
.
Now, plot these two points on to a number line.
Notice that these two numbers effectively divide up the number line into three regions:
, 
, and 
Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.
For 
, let 
Since this 
is not less than 
, the solution to this inequality cannot lie in this region.
For 
, let 
.
Since this will make the inequality true, the solution can lie in this region.
Finally, for 
, let 
Since this number is not less than zero, the solution cannot lie in this region.
Thus, the solution to this inequality is 
Start by setting the inequality to zero and by solving for
Now, plot these two points on to a number line.
Notice that these two numbers effectively divide up the number line into three regions:
Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.
For
Since this
For
Since this will make the inequality true, the solution can lie in this region.
Finally, for
Since this number is not less than zero, the solution cannot lie in this region.
Thus, the solution to this inequality is
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Solve:
Solve:
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First, set the inequality to zero and solve for 
.
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For 
, let 
Since this is not less than zero, the solution to the inequality cannot be found in this region.
For 
, let 
Since this is less than zero, the solution is found in this region.
For 
, let 
Since this is not less than zero, the solution is not found in this region.
Then, the solution for this inequality is 
First, set the inequality to zero and solve for
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For
Since this is not less than zero, the solution to the inequality cannot be found in this region.
For
Since this is less than zero, the solution is found in this region.
For
Since this is not less than zero, the solution is not found in this region.
Then, the solution for this inequality is
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Solve:
Solve:
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First, set the inequality to zero and solve for 
.
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For 
, let 
Since this solution is greater than or equal to 
, the solution can be found in this region.
For 
, let 
Since this is less than or equal to 
, the solution cannot be found in this region.
For 
, let 
Since this is greater than or equal to 
, the solution can be found in this region.
Because the solution can be found in every single region, the answer to this inequality is 
First, set the inequality to zero and solve for
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For
Since this solution is greater than or equal to
For
Since this is less than or equal to
For
Since this is greater than or equal to
Because the solution can be found in every single region, the answer to this inequality is
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Solve:
Solve:
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Start by changing the less than sign to an equal sign and solve for 
.
Now, plot these two numbers on a number line.
Notice how the number line is divided into three regions:
Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.
For 
, let 
Since this number is not less than zero, the solution cannot be found in this region.
For 
, let 
Since this number is less than zero, the solution can be found in this region.
For 
let 
.
Since this number is not less than zero, the solution cannot be found in this region.
Because the solution is only negative in the interval 
, that must be the solution.
Start by changing the less than sign to an equal sign and solve for
Now, plot these two numbers on a number line.
Notice how the number line is divided into three regions:
Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.
For
Since this number is not less than zero, the solution cannot be found in this region.
For
Since this number is less than zero, the solution can be found in this region.
For
Since this number is not less than zero, the solution cannot be found in this region.
Because the solution is only negative in the interval
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Which value for 
would satisfy the inequality 
?
Which value for
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First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: 
. Now we know that the quadratic has zeros at 
and 
. Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this:
We can see that the parabola is below the x-axis (in other words, less than 
) between these two zeros 
and 
.
The only x-value satisfying the inequality 
is 
.
The value of 
would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to 
, that value will not work.
First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us:
We can see that the parabola is below the x-axis (in other words, less than
The only x-value satisfying the inequality
The value of
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Find the roots of the equation x_2 + 5_x + 6 = 0
Find the roots of the equation x_2 + 5_x + 6 = 0
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To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)
So (x + 2)(x + 3) = 0
x = –2 or x = –3
To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)
So (x + 2)(x + 3) = 0
x = –2 or x = –3
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Solve for 
.
Solve for
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First factor the equation. Find two numbers that multiply to 24 and sum to -10. These numbers are -6 and -4: 
Set both expressions equal to 0 and solve for x:
First factor the equation. Find two numbers that multiply to 24 and sum to -10. These numbers are -6 and -4:
Set both expressions equal to 0 and solve for x:
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Solve for 
:
Solve for
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To factor, find two numbers that sum to 5 and multiply to 6.
Check the possible factors of 6:
1 * 6 = 6
1 + 6 = 7, so these don't work.
2 * 3 = 6
2 + 3 = 5, so these work!
Next, pull out the common factors of the first two terms and then the second two terms:
Set both expressions equal to 0 and solve:
and
To factor, find two numbers that sum to 5 and multiply to 6.
Check the possible factors of 6:
1 * 6 = 6
1 + 6 = 7, so these don't work.
2 * 3 = 6
2 + 3 = 5, so these work!
Next, pull out the common factors of the first two terms and then the second two terms:
Set both expressions equal to 0 and solve:
and
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Solve for x.
Solve for x.
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- Split up the middle term so that factoring by grouping is possible.
Factors of 10 include:
1 * 10= 10 1 + 10 = 11
2 * 5 =10 2 + 5 = 7
–2 * –5 = 10 –2 + –5 = –7 Good!
- Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.
- Now pull out the common factor, the "(x-2)," from both terms.
- Set both terms equal to zero to find the possible roots and solve using inverse operations.
x – 5 = 0, x = 5
x – 2 = 0, x = 2
- Split up the middle term so that factoring by grouping is possible.
Factors of 10 include:
1 * 10= 10 1 + 10 = 11
2 * 5 =10 2 + 5 = 7
–2 * –5 = 10 –2 + –5 = –7 Good!
- Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.
- Now pull out the common factor, the "(x-2)," from both terms.
- Set both terms equal to zero to find the possible roots and solve using inverse operations.
x – 5 = 0, x = 5
x – 2 = 0, x = 2
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Solve for x.
Solve for x.
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- First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.
- There are two ways to do this problem. The first and most intuitive method is standard factoring.
16 + 1 = 17
8 + 2 = 10
4 + 4 = 8
- Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.
- Pull out the "(x+4)" to wind up with:
- Set each term equal to zero.
x + 4 = 0, x = –4
But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., 
), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses.
And x, once again, is equal to –4.
- First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.
- There are two ways to do this problem. The first and most intuitive method is standard factoring.
16 + 1 = 17
8 + 2 = 10
4 + 4 = 8
- Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.
- Pull out the "(x+4)" to wind up with:
- Set each term equal to zero.
x + 4 = 0, x = –4
But there's a shortcut! Assuming the terms are arranged by descending degree (i.e.,
And x, once again, is equal to –4.
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Solve for x.
Solve for x.
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- Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.
Then combine like terms.
- Now factor.
1 + 16 = 17
4 + 4 = 8
2 + 8 = 10
- Pull out common factors, "x" and "8," respectively.
- Pull out "(x+2)" from both terms.
x = –8, –2
- Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.
Then combine like terms.
- Now factor.
1 + 16 = 17
4 + 4 = 8
2 + 8 = 10
- Pull out common factors, "x" and "8," respectively.
- Pull out "(x+2)" from both terms.
x = –8, –2
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Solve for x.
Solve for x.
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- Combine like terms and simplify.
No further simplification is possible. The first term has a coefficient that can't be factored away. FOIL requires that all terms be multiplied by each other at some point, so the presence of the coefficient has to be reflected in every step of the factoring.
- Practically speaking, that means we add an extra step. Multiply the coefficient of the first term by the last term before factoring.
3 * 6 = 18
Factors of 18 include:
1 + 18 = 19
2 + 9 = 11
- Now pull out the common factor in each of the pairs, "3x" from the first two and "2" from the second two.
- Pull out the "(x+3)" from both terms.
- Set both parts equal to zero and solve.
3x + 2 = 0, x = –2/3
x + 3 = 0, x = –3
- Combine like terms and simplify.
No further simplification is possible. The first term has a coefficient that can't be factored away. FOIL requires that all terms be multiplied by each other at some point, so the presence of the coefficient has to be reflected in every step of the factoring.
- Practically speaking, that means we add an extra step. Multiply the coefficient of the first term by the last term before factoring.
3 * 6 = 18
Factors of 18 include:
1 + 18 = 19
2 + 9 = 11
- Now pull out the common factor in each of the pairs, "3x" from the first two and "2" from the second two.
- Pull out the "(x+3)" from both terms.
- Set both parts equal to zero and solve.
3x + 2 = 0, x = –2/3
x + 3 = 0, x = –3
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Solve for x.
Solve for x.
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This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get 
.
We then notice that all four numbers are divisible by four, so we can simplify the expression to 
.
Think of the equation in this format to help with the following explanation.
We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.
Since c is negative, we know that our factoring will produce a positive and negative number.
We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.
We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.
Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of 
.
We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case 
.
This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get
We then notice that all four numbers are divisible by four, so we can simplify the expression to
Think of the equation in this format to help with the following explanation.
We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.
Since c is negative, we know that our factoring will produce a positive and negative number.
We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.
We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.
Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of
We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case
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.
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Solve for 
:
Solve for
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To solve for 
, you need to isolate it to one side of the equation. You can subtract the 
from the right to the left. Then you can add the 6 from the right to the left:
Next, you can factor out this quadratic equation to solve for 
. You need to determine which factors of 8 add up to negative 6:
Finally, you set each binomial equal to 0 and solve for 
:
To solve for
Next, you can factor out this quadratic equation to solve for
Finally, you set each binomial equal to 0 and solve for
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