Solving Quadratic Functions - SAT Subject Test in Math I

1. First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

1. 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

1. Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

1. Pull out the "(x+4)" to wind up with:

1. 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.

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

If is one odd number, then the next odd number is . If their product is 143, then the following equation is true.

Distribute into the parenthesis.

Subtract 143 from both sides.

This can be solved by factoring, or by the quadratic equation. We will use the latter.

We are told that both integers are positive, so .

The other integer is .

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 :

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to . We can use this information to find out what and are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

• 3*-1 (sum = 2)
• -3*1 (sum = -2)

Thus our factored equation should look like this:

The roots of the quadratic equation are the values of x for which y is 0.

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

Factor:

Double check by factoring:

Add together:

Therefore:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Your next step would be to take the square root of both sides. At this point, however, you know that you cannot solve the problem. When you take the square root of both sides, you will be forced to take the square root of . This is impossible (at least in terms of real numbers), meaning that this problem must have no real solution.

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}.

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}.

When the baseball hits the ground, the height is 0, so we set . and solve for .

This can be done using the quadratic formula:

Set :

One possible solution:

We throw this out, since time must be positive.

The other:

This solution, we keep. The baseball hits the ground in about 4.7 seconds.

By definition, , so

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfecct square factors of this term are and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

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.

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.

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.