

Algebra, ABSOLUTE Value,
and Exponents
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.
Algebra and Absolute Value
To solve an equation in which the variable is within absolute
value brackets, you have to follow a twostep process:
 Isolate the expression within the absolute value brackets.
 Divide the equation into two.
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:
x + 3 = 5
x + 3 = –5
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 {y^{2} –
3, –y^{2} –1}.
Algebra, Exponents, and Radicals
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: x^{2} =
36. Now you need to take the square root of both sides: .
