This lesson is part 2 of *What to Know for the SAT*. If you haven’t read the first part yet, do so here: What to Know for the SAT (Math-Formulas and Last Minute Tips) [Part 1]

## Radical Equations

On the SAT, you’ll likely encounter a problem involving a radical equation.

### EXAMPLE

√(3x+4) = x

What are all values of *x* that satisfy the equation above?

- A. {-1, 4}
- B. {-1}
- C. {4}
- D. {0, 4}

**SOLUTION**

To solve a problem like this on the SAT, your best bet is to throw away the algebraic way of solving the equation and simply plug in the answer choices.

The answer choices only have 3 different values: -1, 0 and 4. It shouldn’t be too much work plugging in these values. Let’s start with *x* = -1.

Left Hand Side | Right Hand Side | |

√(3x+4) | ? | x |

√(3(-1)+4) | ? | -1 |

√(1) | ? | -1 |

1 | ≠ | -1 |

hand sides of the radical equation

above are not equal at x = -1.

Since the left hand side is NOT equal to the right hand side (1 ≠ -1), we can eliminate answer choices A and B.

For problems of this type, it’s important to remember that when we calculate the square root of a number, we **always take the principal square root** which is the positive square root.

This sometimes gives students difficulty. In the above calculation, when we calculated the square root of 1, the only answer is 1, not -1.

This is really important to remember!!!! Even though it’s true that both 1^{2} = 1 and (-1)^{2} = 1, when we evaluate the square root of a number we ONLY take the positive square root!

Now let’s test the value *x* = 0. I purposely chose this value as the next one to test because 4 is in both answer choices C and D so we know that it is correct.

Left Hand Side | Right Hand Side | |

√(3x+4) | ? | x |

√(3(0)+4) | ? | 0 |

√(4) | ? | 0 |

2 | ≠ | 0 |

hand sides of the radical equation

above are not equal at x = 0.

Since the left hand side does not equal the right hand side, 0 is not a solution. We can eliminate answer choice D.

Therefore, the correct answer is Choice C. (You could verify that *x* = 4 is a solution, though that’s not necessary.)

You may wonder, why didn’t we just square both sides and solve the resulting quadratic equation? Great question!

This is the way you were taught to solve radical equations in math class so why not do that here? The reason has to do with **extraneous solutions**!

Recall that when squaring both sides of an equation, you may introduce extraneous solutions; that is, you may find solutions that satisfy the new equation but do not satisfy the original square root equation.

So, after you work through all the algebra to find the solutions you will still have to take all of your answers and plug them into the original equation to see if they work.

Why not skip all the algebra and just start by plugging in? Time is precious on the SAT so any way you can save time helps tremendously!

## Ratios

Before taking the SAT, make sure you review some ratio problems.

Ratio problems are usually among the easier problems but you certainly don’t want to miss out on points because you neglected to review this concept.

Remember, the ratio of two numbers is simply the quotient of the numbers. Ratios can be written in a few ways: 2 to 3, ⅔, or 2:3 are three different ways to write the ratio 2 to 3.

In general, we want to use fractions to represent ratios.

### EXAMPLE

The ratio of students to teachers at Lincoln High School is 16 to 1. If the school has 68 teachers, how many students does the school have?

**SOLUTION**

To solve this, we can set up a proportion and cross multiply. When working with proportions, it’s essential that you are consistent with how you set up the left side and the right side.

For this problem, on the left side we’ll have the ratio of students to teachers, or 16/1. This means that our ratio on the right side also needs to be students to teachers.

On the right side, let *x* represent the number of students. The denominator is the number of teachers, 68. So we have:

- 16/1 = x/68

Now, we cross multiply:

- 1*x = 16*68
- x = 1,088

Hence, there are 1,088 students at Lincoln High School.

### Equating the Coefficients

On a 2021 SAT exam, students were tasked with solving this problem:

**SOLUTION**

Our best bet in solving this problem is to use a method called **equating the coefficients**. If you have never heard of this method, no problem! We’ll go through it now.

To solve this problem, we’ll work with the left side and right side of the equation separately. We’ll start by multiplying out the left side.

Left Side | Right Side |

x(x+2)^{2} | x^{3} + bx^{2} + cx |

x(x+2)(x+2) | x^{3} + bx^{2} + cx |

x(x^{2}+4x+4) | x^{3} + bx^{2} + cx |

x^{3}+4x^{2}+4x | x^{3} + bx^{2} + cx |

cubic, and we equate coefficients to solve.

Notice, both the left side and the right side expressions are cubic functions. In order for the two cubic functions to be equal, the **coefficients of each term must be equal**.

That is, the coefficients of the cubic terms on both sides must match, the coefficients of the squared terms must match, and so on. In other words, we can **equate the coefficients**!

Set the coefficients equal to each other to determine the values of *b* and *c*. For our problem, since the coefficient of the squared term on the left side is 4, we conclude that *b* = 4.

Since the coefficient of the linear term on the left side is 4, we conclude that *c* = 4. Adding *b* + *c*, we obtain 4 + 4 = 8. The correct answer is Choice C.

## Absolute Value Equations

Be sure you know how to work with absolute value equations because they are easy questions to get correct.

Recall that the absolute value of a number is its distance to the origin on a number line. So, the absolute value of a number is never negative!

**Recall:**

- |5| = 5
- |-3| = 3
- |0| = 0

### EXAMPLE

|2x – 1| = 5

What is the sum of the solutions to the given equation?

**SOLUTION**

To solve an absolute value equation, it’s important to isolate the absolute value part of the equation on one side. In this case, the absolute value is already isolated on the left side.

We know the absolute value of the expression must equal 5 so the expression inside the absolute value brackets must equal 5 or -5. Hence, to find our solutions, we set up two different equations separated by “or.”** **

2x-1=52x=6x=3 | or | 2x-1=-52x=-4x=-2 |

Our two solutions are 3 or -2. The sum of these solutions is 3 + -2 =1 so the answer is 1.

## Unit Conversion

You will often see problems on the SAT that involve unit conversion. For some of the more difficult unit conversion problems, you can use unit analysis to help guide you to the answer.

### EXAMPLE

Two airplanes are traveling at different speeds. The first airplane travels at 620 miles per hour and the second airplane travels at 700 miles per hour.

After 18 seconds, how much farther has the first airplane traveled than the second?

- A) 0.2 miles
- B) 0.4 miles
- C) 0.5 miles
- D) 0.6 miles

**SOLUTION**

We can use the units in the problem to help us find the answer. We’ll consider each airplane separately, then subtract the results. We’ll convert miles per hour to miles per second, then we’ll multiply by 18.

The faster airplane travels 3.5 – 3.1 = 0.4 miles farther than the slower airplane. The correct answer is Choice B.

## Factoring

On the SAT, you need to be confident in your factoring abilities! In the chart below you’ll see the types of factoring problems you’ll likely encounter on the SAT. Review these types before the exam!

Factoring Type | Example | Solution |

Factoring out a Common Monomial | 3x^{2}-6x | 3x(x-2) |

Difference of Squares | 16x^{2}-9 | (4x-3)(4x+3) |

Quadratic Trinomial, leading coefficient equal to 1 | x^{2}-6x+8 | (x-2)(x-4) |

Quadratic Trinomial, leading coefficient not 1 | 3x^{2}+14x-5 | (3x-1)(x+5) |

quadratics, along with solutions.

### EXAMPLE

If x+y=3 and x-y=5, what is the value of (x^{2}-y^{2})(x+y)?

SOLUTION

At first glance, this problem looks difficult! It seems as though it involves a system of equations as well as some kind of factoring expression.

The long way to do this problem would be to solve the system of linear equations, then plug in the values of *x* and *y* to the expression (x^{2}-y^{2})(x+y). This method will work but will take a fair amount of time.

A faster approach: factor (x^{2}-y^{2})(x+y).

Notice, the expression (x^{2}-y^{2}) is the special case of factoring, the difference of squares. The SAT loves the difference of squares!!!

The expression (x^{2}-y^{2}) factors to (x-y)(x+y). So we obtain:

- (x-y)(x+y)(x+y)

Now, here’s the interesting part. We already know from the given that x+y=3 and x-y=5. We can actually plug in 3 for the expression (x+y) and 5 for the expression (x-y)!

- (x-y)(x+y)(x+y) =5*3*3 = 45

So, the correct answer is 45.

### EXAMPLE

Find the *x*-intercepts of the parabola y = x^{2}-6x-7.

**SOLUTION**

To find the *x*-intercepts of a parabola, we can set the equation equal to 0, factor, and solve.

To factor a quadratic trinomial with the leading coefficient of 1, we simply look for the pair of numbers that will multiply to the constant -7 that will also combine to -6.

The only possibility is -7 and 1 because the product -71=-1 and the sum is -7+1=-6.

- x
^{2}-6x-7=0 - (x-7)(x+1) = 0
- x-7=0 or x+1=0
- x=7 or x=-1

The two *x*-intercepts are *x* = 7 and *x* = -1.

Hopefully, this lesson helped you with your SAT Preparation! If you think you’re ready, a great next step is to start taking some practice tests.

Everything you need is on the College Board’s website: SAT Practice Tests.

Be aware: the practice tests are long! Each one is about 65 pages so if you decide to print an exam, you’ll use a lot of paper and a lot of ink! You could also purchase the College Board’s workbook with practice tests.

I hope you found this article helpful. If so, please share it with someone who can use the information.

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**About the author:**

Jean-Marie Gard is an independent math teacher and tutor based in Massachusetts. You can get in touch with Jean-Marie at https://testpreptoday.com/.