All ACT Math Resources
Example Questions
Example Question #1 : Quadratic Equations
In the equation what are the values of ?
None of the other answers
To solve, begin by factoring the equation. We know that in our two factors, one will begin with and one will begin with . Fiddling with different factors of (we are looking for two numbers that, when multiplied by and separately, will add to ), we come to the following:
(If you are unsure, double check by expanding the equation to match the original)
Now, set the each factor equal to 0:
Do the same for the second factor:
Therefore, our two values of are and .
Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.
Example Question #1 : Quadratic Equations
Solve: x2+6x+9=0
6
-3
3
9
12
-3
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.
Example Question #1 : 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 : Quadratic Equations
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 #2 : Quadratic Equations
36x2 -12x - 15 = 0
Solve for x
-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
-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 : How To Factor The 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 #2035 : Act Math
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 #11 : Quadratic Equations
What is the sum of the two solutions of the equation x2 + 5x – 24 = 0?
3
-5
24/5
-24/5
-24
-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 #12 : Quadratic Equations
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
8
6
2
10
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 #11 : 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.