Equations / Inequalities - ACT Math

Card 0 of 20

Question

If (x2 + 2) / 2 = (x2 - 6x - 1) / 5, then what is the value of x?

Answer

(x2 + 2) / 2 = (x2 - 6x - 1) / 5. We first cross-multiply to get rid of the denominators on both sides.

5(x2 + 2) = 2(x2 - 6x - 1)

5x2 + 10 = 2x2 - 12x - 2 (Subtract 2x2, and add 12x and 2 to both sides.)

3x2 + 12x + 12 = 0 (Factor out 3 from the left side of the equation.)

3(x2 + 4x + 4) = 0 (Factor the equation, knowing that 2 + 2 = 4 and 2*2 = 4.)

3(x + 2)(x + 2) = 0

x + 2 = 0

x = -2

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Question

Solve 8x2 – 2x – 15 = 0

Answer

The equation is in standard form, so a = 8, b = -2, and c = -15. We are looking for two factors that multiply to ac or -120 and add to b or -2. The two factors are -12 and 10.

So you get (2x -3)(4x +5) = 0. Set each factor equal to zero and solve.

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Question

If , and , which of the following is a possible value of ?

Answer

The given expression is a quadratic equation; therefore, we can factor the equation

Use the format of the standard quadratic equation:

Since , we know that the quadratic's roots will resemble the following:

$(m + \ )(m -\ ) = 0$

We also know that one of those signs has to be negative, since our two last terms multiply to equal the variable , and is negative in our quadratic. Now, we need to find two numbers that when multiplied together equal -24, and equal 10 when they are added together. Let's start by finding the factors of 24. The factors of 24 are 24 and 1, 12 and 2, 8 and 3, 6 and 4. Since one of those factors will be negative in our factored equation, we need to find the two factors whose difference is 10.

This means that the numbers in the factored equation are 12 and -2; thus, we may write the following:

.

By the zero multiplication rule, either portion of that equation can equal 0 for the result to be 0; thus, we have the following two expressions:

Subtract 12 from both sides of the equation:

Calculate the value of the variable in the second equation.

Add 2 to both sides of the equation.

Since we want a negative answer for our variable, the correct answer is:

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Question

What is the value of where:

Answer

The question asks us to find the value of , because it is in a closed equation, we can simply put all of the whole numbers on one side of the equation, and all of the containing numbers on the other side.

We utilize opposite operations to both sides by adding to each side of the equation and get

Next, we subtract from both sides, yielding

Then we divide both sides by to get rid of that on

$x=\frac{1}{5}$

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Question

For what value of b is the equation b2 + 6b + 9 = 0 true?

Answer

Factoring leads to (b+3)(b+3)=0. Therefore, solving for b leads to -3.

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Question

What is the solution to:

$\frac{x^2-6x+8}{x-2}=0$

Answer

First you want to factor the numerator from x2 – 6x + 8 to (x – 4)(x – 2)

Input the denominator (x – 4)(x – 2)/(x – 2) = (x – 4) = 0, so x = 4.

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Question

Which of the following is a factor of the polynomial x_2 – 6_x + 5?

Answer

Factor the polynomial by choosing values that when FOIL'ed will add to equal the middle coefficient, 3, and multiply to equal the constant, 1.

x_2 – 6_x + 5 = (x – 1)(x – 5)

Because only (x – 5) is one of the choices listed, we choose it.

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Question

Factor the following equation:

${x^3-2x^2+x}$

Answer

First we factor out an x then we can factor the

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Question

7 times a number is 30 less than that same number squared. What is one possible value of the number?

Answer

$\small 7x+30=x^{2}$

$\small x^{2}-7x-30=0$

$\small (x-10)(x+3)=0$

Either:

$\small x-10=0$

$\small x=10$

or:

$\small x+3=0$

$\small x=-3$

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Question

Which of the following is equivalent to $2x^{2}\left ( xy^{2}+5x^{2}y^{2} \right )$ ?

Answer

The answer is $2x^{3}y^{2}+10x^{4}y^{2}$ .

To determine the answer, $2x^{2}$ must be distrbuted,

$(2x^{2}xy^{2}) +(2x^{2}5x^{2}y^{2})$ . After multiplying the terms, the expression simplifies to $2x^{3}y^{2}+10x^{4}y^{2}$ .

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Question

Which of the following equations is NOT equivalent to the following equation?

$4y^{2}=169x^{2}-81$

Answer

The equation presented in the problem is:

$4y^{2}=169x^{2}-81$

We know that:

$169x^{2}-81=(13x)^{2}-9^{2}=(13x+9)(13x-9)$

Therefore we can see that the answer choice $(2y)^{2}=(13x+9)(13x-9)$ is equivalent to $4y^{2}=169x^{2}-81$ .

$\frac{(2y)^{2}}{13x+9}=27x-9-14x$ is equivalent to $4y^{2}=169x^{2}-81$ . You can see this by first combining like terms on the right side of the equation:

$\frac{(2y)^{2}}{13x+9}=13x-9$

Multiplying everything by , we get back to:

$(2y)^{2}=(13x+9)(13x-9)$

We know from our previous work that this is equivalent to $4y^{2}=169x^{2}-81$ .

$6y^{2}=\frac{3\times (13x+9)\times (13x-9)}{2}$ is also equivalent $4y^{2}=169x^{2}-81$ since both sides were just multiplied by $\frac{3}{2}$ . Dividing both sides by $\frac{3}{2}$ , we also get back to:

$(2y)^{2}=(13x+9)(13x-9)$ .

We know from our previous work that this is equivalent to $4y^{2}=169x^{2}-81$ .

$y=\sqrt{\frac{338x^{2}-162}{8}}$ is also equivalent to $4y^{2}=169x^{2}-81$ since

$y^{2}=(\sqrt{\frac{338x^{2}-162}{8}})^{2}=\frac{338x^{2}-162}{8}$

$4y^{2}=\frac{338x^{2}-162}{2}=169x^{2}-81$

Only $y^{2}=(\frac{13x-9}{2})^{2}$ is NOT equivalent to $4y^{2}=169x^{2}-81$

because

$y^{2}=(\frac{13x-9}{2})^{2}=\frac{169x^{2}-234x-81}{4}$

$4y^{2}=169x^{2}-81-234x eq 4y^{2}=169x^{2}-81$

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Question

A certain number squared, plus four times itself is equal to zero.

Which of the following could be that number?

Answer

The sentence should first be translated into an equation.

We will call the "certain number" .

This gives us $x^{2}+4x=0$ .

So solve this we can factor out an to get .

This makes the zeros of the equation evident. We know that when the entire expression will equal zero, making the equation true. We also know that when , the quantity will be , which also satisfies the equation.

Therefore, the two possible solutions are and .

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Question

Find the roots of the equation .

Answer

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)

(x + 2)(x + 3) = 0

x = –2 or x = –3

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Question

If x2 + x – 6 = 0, what does x equal?

Answer

There are two ways to solve this problem. One option is to use the quadratic formula:

In this problem, a = 1, b = 1, and c = –6. However, there is a lot of room for computational error using this method.

Another option is to factor the equation using the reverse FOIL method. Because we have x2, we know that both factors must include x. Thus, so far we have (x )(x ). We also know that one factor must be negative and one must be positive if our last value (the L in FOIL) is to be negative. Thus, we have (x– ) (x+ ). The factors of 6 are 1, 2, 3, and 6, so our last values must multiply together to make 6. Therefore, our options are (x – 1)(x + 6), (x – 6)(x + 1), (x – 2)(x + 3), and (x – 3)(x + 2). However, because we need our middle value to be 1, the combination of the outer and inner values (O and I in FOIL), must add up to positive 1. Since only a combination of +/–2 and +/–3 can add to 1, we can eliminate our first two options. Then, because we need the positive value to be larger than the negative value, we can determine that the correct factorization is (x – 2)(x + 3). When we set each of these pieces equal to zero, we get x – 2 = 0 and x + 3 = 0. Thus, X equals 2 and –3. We can check our work using FOIL.

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Question

Solve: x2+6x+9=0

Answer

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.

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Question

36x2 -12x - 15 = 0

Solve for x

Answer

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

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Question

64x2 + 24x - 10 = 0

Solve for x

Answer

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

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Question

x_2 + 3_x + 2 = 0

What are the values of x that are solutions to the equation?

Answer

You factor the equation and get:

(x + 2)(x + 1) = 0

Then, you get x = –2 and x = –1.

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Question

Which of the following is a root of the function $f(x)=2x^2-7x-4$ ?

Answer

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

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Question

Which of the following values of are solutions to $x^{2}-3x-28=0$ ?

Answer

1. Factor the above equation:

$x^{2}-3x-28=0$

2. Solve for :

and...

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