All ACT Math Resources
Example Questions
Example Question #2 : Quadratic Equations
64x2 + 24x - 10 = 0
Solve for x
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
Example Question #1 : How To Factor The Quadratic Equation
Which of the following is a root of the function ?
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 2x from the first two terms.
Thus, when factored, the original equation becomes (2x + 1)(x – 4) = 0.
We now set each factor equal to zero and solve for x.
Subtract 1 from both sides.
2x = –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 .
Example Question #5 : Quadratic Equations
36x2 -12x - 15 = 0
Solve for x
1/2 and 5/6
-1/2 and 5/6
1/2 and 1/3
1/2 and -1/3
-1/2 and -5/6
-1/2 and 5/6
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
Example Question #3 : Quadratic Equation
What is the sum of all the values of that satisfy:
With quadratic equations, always begin by getting it into standard form:
Therefore, take our equation:
And rewrite it as:
You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:
Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :
OR
The sum of these values is:
Example Question #11 : Quadratic Equations
What are the two solutions to the following quadratic equation?
and
and
and
and
and
and
The equation can be factored.
First pull out a common factor of three from each term.
.
Now, find the factors of the constant term that when added together result in the middle term.
The roots are what values will make the equation equal .
Therefore, the answers are and
Example Question #541 : Algebra
What is the sum of the two solutions of the equation x2 + 5x – 24 = 0?
24/5
-24
3
-24/5
-5
-5
First you must find the solutions to the equation. This can be done either by using the quadratic formula or by simply finding two numbers whose sum is 5 and whose product is -24 and factoring the equation into (x + 8)(x – 3) = 0. The solutions to the equation are therefore -8 and 3, giving a sum of -5.
Example Question #542 : Algebra
The height of a ball (in feet) after it is thrown in the air is given by the expression
s(t) = –t2 + 4t
where t is time in seconds. The ball is thrown from ground level at t = 0. How many seconds will pass before the ball reaches the ground again?
4
10
6
2
8
4
Notice that when the ball is at ground level, the height is zero. Setting s (t) equal to zero and solving for t will then give the times when the ball is at the ground.
–t2 + 4t =0
t(4 – t) = 0
t = 0, t = 4
The ball returns to the ground after 4 seconds.
Example Question #12 : Quadratic Equations
Two positive consecutive multiples of 3 have a product of 180. What is the sum of the two numbers?
Define varables as = the first number and = the second number. The product of the numbers is . Solve the resulting quadratic equation by factoring and setting each factor to zero. The numbers are 12 and 15 and the sum is 27.
Example Question #13 : Quadratic Equations
Given the equation: .
What is the product of the solutions of the quadratic equation?
We are initially presented with a quadratic equation, . To begin we must factor this equation.
The multiples of 15 are (15 and 1) and (3 and 5). The only multiples that add or subtract to are 3 and 5. Hence we use these as our binomial numbers. . We must now decide on the signs. Because we need to add or subtract 5 and 3 to get to , both signs must be negative: .
From this point we need to switch gears to find solutions to the equation. What numbers would make this equation equal 0?
At this point split the equation into two parts.
and and solve.
and . Both of these numbers inserted into the original equation will produce a result of 0.
Now the question itself is asking for the product of the solutions to the equation, or , which equals 15, therefore 15 is our answer.
Example Question #14 : Quadratic Equations
If and , what is the greatest value that can have?
Solving for yields and. Solving for yields and .
The greatest difference between these two numbers is 14, and 14 squared is 196.
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