Equations / Inequalities - PSAT Math

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.

36x2 - 12x - 15 = 0

Factor the equation:

(6x + 3)(6x - 5) = 0

Set each side equal to zero

6x + 3 = 0

x = -3/6 = -1/2

6x – 5 = 0

x = 5/6

64x2 + 24x - 10 = 0

Factor the equation:

(8x + 5)(8x – 2) = 0

Set each side equal to zero

(8x + 5) = 0

x = -5/8

(8x – 2) = 0

x = 2/8 = 1/4

The roots of an equation are the points at which the function equals zero. We can set each function equal to zero and determine which functions have one root, and which does not.

Another piece of information will help. If a quadratic function has one root, then it must be a perfect square. This is because a quadratic function that is a perfect square can be written in the form (x – a)2. If we set (x – a)2 = 0 in order to find the root, we see that a is the only value that solves the equation, and thus a is the only root. Additionally, a quadratic equation is a perfect square if it can be written in the form a2x2 + 2abx + b2 = (ax + b)2.

Let's examine the choice f(x) = 4x2 – 4x+1. To find the roots, we set f(x) = 0.

4x2 – 4x+1 = 0

We notice that 4x2 - 4x + 1 is a perfect square, since we could write it as (2x – 1)2. Thus, this equation has only one root, and it can't be the answer.

If we look at f(x) = x2 –2x + 1, we see that x2 – 2x + 1 is also a perfect square, because it could be written as (x – 1)2. This function also has a single root.

Next, we examine f(x) = (1/4)x2 + x + 1. Let us set f(x) = (1/4)x2 + x + 1 = 0.

(1/4)x2 + x + 1 = 0

We can multiply both sides by four to get rid of the fraction.

x2 + 4x + 4 = 0

(x + 2)2 = 0

This function is also a perfect square and has a single root.

Now consider the choice f(x) = (–1/9)x2 + 6x – 81.

f(x) = (–1/9)x2 + 6x – 81 = 0

Multiply both sides by –9.

x2 – 54x + 729 = 0

(x – 27)2 = 0.

Finally, let's look at f(x) = 9x2 – 6x + 4. This CANNOT be written as a perfect square, because it is not in the form a2x2 + 2abx + b2 = (ax + b)2. It might be tempting to think that 9x2 - 6x + 4 = (3x - 2)2, but it does NOT, because (3x – 2)2 = 9x2 – 12x + 4. Therefore, because 9x2 – 6x + 4 is not a perfect square, it doesn't have exactly one root.

x2 – x = 72. Solve for x using the quadratic formula and x = 9 and –8. Only 9 satisfies the restrictions.

The roots of a function are the x intercepts of the function. Whenever a function passes through a point on the x-axis, the value of the function is zero. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values of x.

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

Because the coefficient in front of the is not equal to 1, we need to multiply this coefficient by the constant, which is –4. When we mutiply 2 and –4, we get –8. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of the x term). Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We will now rewrite –7x as –8x + x.

We will then group the first two terms and the last two terms.

We will next factor out a 2_x_ from the first two terms.

Thus, when factored, the original equation becomes (2_x_ + 1)(x – 4) = 0.

We now set each factor equal to zero and solve for x.

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

Now, we set x – 4 equal to 0.

x – 4 = 0

Add 4 to both sides.

x = 4

The roots of f(x) occur where x = .

The answer is therefore .

Let = number of dimes, = number of nickels, and

= number of quarters.

The general equation to use is:

where is the money value and is the number of coins

So the equation to solve becomes

Thus, solving the equation shows that she had five nickels, four dimes, and three quarters giving a total of 12 coins.

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (3, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (4, or b in the standard quadratic formula). Because their product is positive (3) and the sum is positive, that must mean that they both have positive signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 3, as the product of 1 and 3 is 3, and sum of 1 and 3 is 4. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore, the zeros are,

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (20, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (9, or b in the standard quadratic formula). Because their product is positive (20) and the sum is positive, that must mean that they both have positive signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and 5, as the product of 4 and 5 is 20, and sum of 4 and 5 is 9. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros are,

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (3, or b in the standard quadratic formula). Because their product is negative (-4) and the sum is positive, that must mean that they have opposite signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and -1, as the product of 4 and -1 is -4, and sum of 4 and -1 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-1, or b in the standard quadratic formula). Because their product is negative (-2) and the sum is negative, that must mean that they have opposite signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -2, as the product of 1 and -1 is -2, and sum of -2 and 1 is -1. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore, the zeros of the function are

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square.

When simplified the new function is,

Since the term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the value into the original function.

Therefore the maximum value occurs at the point .

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-9, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-9) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 3 and -3, as the product of 3 and -3 is -9, and sum of 3 and -3 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

and

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (16, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (8, or b in the standard quadratic formula). Because their product is positive (16) and the sum is positive, that must mean that they both have positive signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and 4, as the product of 4 and 4 is 16, and sum of 4 and 4 is 8. So, this results in the expression's factored form looking like...

From here, set the binomial equal to zero and solve for .

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zero of the function is,

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square.

When simplified the new function is,

Since the term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the value into the original function.

Therefore the maximum value occurs at the point .

To find the zeros of this function first identify and factor of the GCF.

In this particular case,the GCF is as it appears in both terms. Factoring out the GCF results in the following.

From here, set each term equal to zero and solve for .

and

Therefore the zeros are,

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square.

When simplified the new function is,

Since the term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the value into the original function.

Therefore the maximum value occurs at the point .

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

When simplified the new function is,

Since the term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the value into the original function.

Therefore the minimum value occurs at the point .