How to factor the quadratic equation - 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.

$f(x)=2x^2$-7x-4 = 0

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 $x^2$ 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.

$2x^2$$-7x-4=2x^2$-8x+x-4=0

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

$(2x^2$-8x)+(x-4)=0

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

$(2x^2$-8x)+(x-4)=2x(x-4)+1(x-4)=(2x+1)(x-4)=0

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.

2x + 1 = 0

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

x=-$\frac{1}{2}$

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 = -$\frac{1}{2}$,4.

The answer is therefore x = -frac {1}{2}.

We can factor the quadratic equation into $(x-3)^{2}$=0.

Then we can see that x=3.

Therefore, xcdot k=1 becomes 3cdot k=1 and k=\frac{1}{3}$.

Since you have an addition for the last element, you know that you will have either two subtractions or two additions in your factored groups. Since the second term is positive, your factored groups will contain two additions:

Now, all you have to do are go through the factors of 30 and find the pair that adds up to 11:

and : No!

and : No!

and : No! (But closer!)

and : Yes!

Therefore, your answer will be:

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.

$f(x)=2x^2$-7x-4 = 0

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 $x^2$ 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.

$2x^2$$-7x-4=2x^2$-8x+x-4=0

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

$(2x^2$-8x)+(x-4)=0

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

$(2x^2$-8x)+(x-4)=2x(x-4)+1(x-4)=(2x+1)(x-4)=0

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.

2x + 1 = 0

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

x=-$\frac{1}{2}$

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 = -$\frac{1}{2}$,4.

The answer is therefore x = -frac {1}{2}.

We can factor the quadratic equation into $(x-3)^{2}$=0.

Then we can see that x=3.

Therefore, xcdot k=1 becomes 3cdot k=1 and k=\frac{1}{3}$.

Since you have an addition for the last element, you know that you will have either two subtractions or two additions in your factored groups. Since the second term is positive, your factored groups will contain two additions:

Now, all you have to do are go through the factors of 30 and find the pair that adds up to 11:

and : No!

and : No!

and : No! (But closer!)

and : Yes!

Therefore, your answer will be:

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