The new SAT puts more emphasis on subjects from Algebra II. In part, this means that the test asks more algebra questions that include absolute value, radicals, and exponents. All three mathematical concepts add certain complications to solving algebra equations.

To solve an equation in which the variable is within absolute value brackets, you have to follow a two-step process:

Divide the equation in two? What? Watch:

Since x + 3 has absolute value brackets around it, for the expression to equal 5, the expresion x + 3 when outside of the absolute value brackets can equal either +5 or –5. So you’re actually dealing with two equations:

To solve the problem, you need to solve both of them. First, solve for x in the equation x + 3 = 5. In this case, x = 2. Then, solve for x in the equation x + 3 = –5. In this case, x = –8. So the solutions to the equation |x + 3| = 5 are x = {–8, 2}.

Here’s another example with a much more complicated equation:

First, isolate the expression within the absolute value brackets:

Remember that in terms of PEMDAS, absolute value brackets are like parentheses—do the math inside them first. So solve for the variable as if the expression within absolute value brackets were positive:

Multiply both sides of the equation by 3:

Subtract 2 from both sides:

Next, solve for the variable as if the expression within absolute value brackets were negative:

Multiply both sides of the equation by 3:

Distribute the negative sign (crucial step, make sure you do this or you’ll fall into a trap!):

Subtract 2 from both sides:

The solution set for x is {y² – 3, –y² –1}.

Exponents and radicals can have devilish effects on algebraic equations that are similar to those caused by absolute value.

Consider the equation . Seems pretty simple, right? Just take the square root of both sides and you end up with x = 5. But remember the rule of multiplying negative numbers?

When two negative numbers are multiplied together the result is a positive. In other words, –5 squared also results in 25: .

This means that whenever you have to take the square root to simplify a variable brought to the second power, the result will be two solutions, one positive and one negative:. The only exception is if x = 0.

Want an example?

To solve this problem, you first simplify the problem by dividing 2 out of both sides: x² = 36. Now you need to take the square root of both sides: .