Tag Archive for: rational expressions

Rationalizing a Denominator

April 2, 2021/in Accuplacer, Algebra, Math Test Prep/by Jill Hacker

This post about rationalizing a denominator is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

Question

Consider the expression

${\color{Blue} \frac{a-b}{\sqrt{a}-\sqrt{b}} }$

where a and b are positive integers.

Which of the following expressions is equivalent to it?

A. $\frac{a-b}{\sqrt{a-b}}$

B. $\sqrt{a-b}$

C. $\sqrt{a}+\sqrt{b}$

D. $a\sqrt{a}+b\sqrt{b}$

Solution

Only one of the four answer choices is a fraction. See if you can convert the expression that you are given into a format that is not a fraction and perhaps get one of the answer choices. Is there a way to make that denominator go away?

Sidebar: Rationalizing a denominator using a conjugate

A denominator that contains radicals is considered poor form. You may sometimes be asked to rationalize such denominators, meaning convert the fraction or rational expression into a format that has no radicals in its denominator. The usual way to do that is to use the denominator’s conjugate. The conjugate of an expression $c+d$ is $c-d$ .

For $\frac{a-b}{\sqrt{a}-\sqrt{b}}$ , multiply and divide by the denominator’s conjugate to rationalize the denominator, which is likely a step in the right direction. (To multiply an expression by something and divide by the same thing changes the expression’s form but not its value.) Let’s try that.

The conjugate of

${\color{Blue} \sqrt{a}-\sqrt{b}}$

${\color{Blue} \sqrt{a}+\sqrt{b}}$

Multiplying and dividing $\frac{a-b}{\sqrt{a}-\sqrt{b}}$ by the conjugate looks like:

${\color{Blue} \frac{a-b}{\sqrt{a}-\sqrt{b}} \cdot \frac {\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left ( a-b \right )\left ( \sqrt{a}+\sqrt{b} \right )}{a-b}}$

All the radicals have dropped out of the denominator. That’s one thing we were looking for. Still, we can simplify: Notice that $a-b$ is a factor in both the numerator and the denominator. Cancel it and you get

${\color{Blue} \frac{\left ( a-b \right )\left ( \sqrt{a}+\sqrt{b} \right )}{a-b}=\sqrt{a}+\sqrt{b}}$

This is answer choice C. So you’re done.

For more about rational numbers, expressions, and equations, see Solving a Rational Equation.

The Accuplacer sample problem on which this one is based is #19 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

Solving a Rational Equation

March 30, 2021/in Accuplacer, Algebra, Math Test Prep/by Jill Hacker

This post about solving a rational equation is #15 in a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

Question

Solve this rational equation:

${\color{Blue} \frac{5}{x+8}=\frac{x}{3x-2}}$

Assume x is not equal to -8 or 2/3.

Solution

What do we mean by rational? The root is ratio. A rational number is a number that can be expressed as quotient — a ratio — of two integers. A rational expression is a quotient — a ratio — of algebraic expressions. A rational equation is an equation comprised of rational expressions.

To get started solving this rational equation, multiply both sides by $\left (x+8 \right )\left (3x+2 \right )$ . That should give you:

${\color{Blue} \frac{5\cdot \left (x+8 \right )\left (3x+2 \right )}{x+8}=\frac{x\left (x+8 \right )\left (3x+2 \right )}{3x-2}}$

Cancel the terms that show up in both numerator and denominator. That’s $x+8$ on the left and $3x-2$ on the right. That leaves:

${\color{Blue} 5 \cdot \left (3x-2 \right )=x\left (x+8 \right )}$

(Yes, I could have just said “cross multiply,” but I hate that expression. It’s a personal quirk.) Now we’re done with denominators.

Distribute:

${\color{Blue} 15x-10=x^2+8x}$

Subtract $15x-10$ from both sides, and then switch the sides.

${\color{Blue} 0=x^2-7x+10}$

${\color{Blue} x^2-7x+10=0}$

Factor:

${\color{Blue} x^2-7x+10=\left (x-2 \right )\left ( x-5 \right )}$

For more about factoring, see “How to Factor a Quadratic.”

Set each factor equal to 0:

$x-2=0$	$x-5=0$
$x=2$	$x=5$

Caution

Remember that division by zero yields something undefined, so an x-value that makes a denominator zero is not a valid answer. So you need to make sure that the x values you’ve found do not make one of the denominators in the original question zero. To find those values, set those denominators equal to zero:

$x+8=0$	$3x-2=0$
$x=-8$	$x=2/3$

Oh, look: Those are the values that the question says are excluded. The question is telling you that a solution that causes the denominator to be zero is not a valid solution. But today that is not your problem, because that’s not one of the answers we found. We found $x=2$ and $x=5$ . Interesting factoid: The official practice question on which this question is based stipulates that x is not equal to -3/2 and then offers -3/2 as an answer choice. If there’s a lesson in that, it may be that it’s a good idea to reread the question before you give your final answer.

To complete the check, substitute $x=2$ and $x=5$ into the original equation.

This question is similar to question number 15 in the sample questions for the Accuplacer Advanced Algebra and Functions test.