All SAT Math Resources
Example Questions
Example Question #3 : Quadratic Equations
I. real
II. rational
III. distinct
Which of the descriptions characterizes the solutions of the equation 2x2 – 6x + 3 = 0?
I and II only
II and III only
II only
I only
I and III only
I and III only
The equation in the problem is quadratic, so we can use the quadratic formula to solve it. If an equation is in the form ax2 + bx + c = 0, where a, b, and c are constants, then the quadratic formula, given below, gives us the solutions of x.
In this particular problem, a = 2, b = –6, and c = 3.
The value under the square-root, b2 – 4ac, is called the discriminant, and it gives us important information about the nature of the solutions of a quadratic equation.
If the discriminant is less than zero, then the roots are not real, because we would be forced to take the square root of a negative number, which yields an imaginary result. The discriminant of the equation we are given is (–6)2 – 4(2)(3) = 36 – 24 = 12 > 0. Because the discriminant is not negative, the solutions to the equation will be real. Thus, option I is correct.
The discriminant can also tell us whether the solutions of an equation are rational or not. If we take the square root of the discriminant and get a rational number, then the solutions of the equation must be rational. In this problem, we would need to take the square root of 12. However, 12 is not a perfect square, so taking its square root would produce an irrational number. Therefore, the solutions to the equation in the problem cannot be rational. This means that choice II is incorrect.
Lastly, the discriminant tells us if the roots to an equation are distinct (different from one another). If the discriminant is equal to zero, then the solutions of x become (–b + 0)/2a and (–b – 0)/2a, because the square root of zero is 0. Notice that (–b + 0)/2a is the same as (–b – 0)/2a. Thus, if the discriminant is zero, then the roots of the equation are the same, i.e. indistinct. In this particular problem, the discriminant = 12, which doesn't equal zero. This means that the two roots will be different, i.e. distinct. Therefore, choice III applies.
The answer is choices I and III only.
Example Question #12 : Quadratic Equations
Solve for x.
3x2 + 15x – 18 = 0.
x = 5 or x = 1
x = –2 or x = 3
x = 6 or x = –1
x = –6 or x = 1
x = 2 or x = –3
x = –6 or x = 1
First let's see if there is a common term.
3x2 + 15x – 18 = 0
We can pull out a 3: 3(x2 + 5x – 6) = 0
Divide both sides by 3: x2 + 5x – 6 = 0
We need two numbers that sum to 5 and multiply to –6. 6 and –1 work.
(x + 6)(x – 1) = 0
x = –6 or x = 1
Example Question #321 : Equations / Inequalities
The expression is equal to 0 when and
Factor the expression and set each factor equal to 0:
Example Question #12 : How To Find The Solution To A Quadratic Equation
Two positive consecutive multiples of four have a product of 96. What is the sum of the two numbers?
Let = the first number and = the second number.
So the equation to solve becomes . This quadratic equation needs to be multiplied out and set equal to zero before factoring. Then set each factor equal to zero and solve. Only positive numbers are correct, so the answer is .
Example Question #21 : How To Find The Solution To A Quadratic Equation A1
Two consecutive positive odd numbers have a product of 35. What is the sum of the two numbers?
Let = first positive number and = second positive number.
The equation to solve becomes
We multiply out this quadratic equation and set it equal to 0, then factor.
Example Question #14 : How To Find The Solution To A Quadratic Equation
The product of two consecutive positive multiples of four is 192. What is the sum of the two numbers?
Let = the first positive number and = the second positive number
The equation to solve becomes
We solve this quadratic equation by multiplying it out and setting it equal to 0. The next step is to factor.
Example Question #22 : Quadratic Equations
Two consecutive positive multiples of three have a product of 504. What is the sum of the two numbers?
Let = the first positive number and = the second positive number.
So the equation to solve is
We multiply out the equation and set it equal to zero before factoring.
thus the two numbers are 21 and 24 for a sum of 45.
Example Question #22 : Quadratic Equations
Two consecutive positive numbers have a product of 420. What is the sum of the two numbers?
Let = first positive number and = second positive number
So the equation to solve becomes
Using the distributive property, multiply out the equation and then set it equal to 0. Next factor to solve the quadratic.
Example Question #23 : Quadratic Equations
A rectangle has a perimeter of and an area of What is the difference between the length and width?
For a rectangle, and where = width and = length.
So we get two equations with two unknowns:
Making a substitution we get
Solving the quadratic equation we get or .
The difference is .
Example Question #182 : Gre Quantitative Reasoning
If then which of the following is a possible value for ?
Since , .
Thus
Of these two, only 4 is a possible answer.