### Tag Archive for: radical equations

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

## Question

Does the equation $\inline \fn_jvn \sqrt{4x+3}+10=5$ have a real solution? If so, what is it?

## Solution

This kind of equation is called a radical equation, because it contains a radical — in this case, a square root.

Let’s try to solve this radical equation:

$\fn_jvn {\color{Blue} \sqrt{4x+3}+10=5}$

Subtract 5 from each side:

$\fn_jvn {\color{Blue} \sqrt{4x+3}=-5}$

The square root of something equals something negative? Really? The definition of square root specifies that it means a positive number. But this square root is supposed to equal something negative.

No real solution.

You may be tempted to keep going from$\fn_jvn {\color{Blue} \sqrt{4x+3}=-5}$

and see what happens. OK, let’s try that. Follow the usual steps for solving a radical equation:

• Isolate the radical. We did that above when we added 10 to each side.
• Square both sides:

$\fn_jvn {\color{Blue} 4x+3=25}$

• Subtract 3 from each side:

$\fn_jvn {\color{Blue} 4x=22}$

• Divide by 4:

$\fn_jvn {\color{Blue} x=11/2}$

There’s a solution: $\inline \fn_jvn x=11/2$. Now let’s see if it works. Substitute $\inline \fn_jvn x=11/2$ back into the left-hand side of the original equation. You should get:

$\fn_jvn {\color{Blue} \sqrt{4 \cdot 11/2+3}+10}$

That simplifies to:

$\fn_jvn {\color{Blue} \sqrt{25}+10= 15\neq5 }$

So what looked like the wrong way was indeed the wrong way, the solution $\inline \fn_jvn x=11/2$ does not work, and the answer to the question is No, the equation $\inline \fn_jvn \sqrt{4x+3}+10=5$ does not have a real solution.

This question is similar to question number 14 in the sample questions for the Accuplacer Advanced Algebra and Functions test.