SAT Practice Question
\(2\sqrt{9x}-6=-10-2\sqrt{x}\) What value of \(x\) is the solution to the above equation? Source: Khan Academy |
Now that you have learned how to work with rational exponents and radicals, we can continue that learning into solving equations that involve these. As we work through these equations, we must be careful to check for extraneous solutions. Recall that extraneous solutions are ones that you get when you solve but don’t work in the original equation. You have seen these before in absolute value equations because when we changed the structure of the absolute value equations, we sometimes got additional solutions that did not work in the original absolute value equation. You will also see extraneous solutions in the next unit as well. We can identify extraneous solutions by substituting the values back into the original equation and seeing if they still produce a true statement.
A quick note about signs of solutions that tend to confuse students. When we see an equation like \(x^2=25\), we can correctly say that the solutions of that equation are \(x=\pm5\) since squaring a positive or negative \(5\) would give us \(25\). However, the equation \(x=\sqrt{25}\) only gives a solution of \(x=5\) since there is just a positive sign in front of the square root. Some of the extraneous solutions we will get below come into play when you end up with something like \(-5=\sqrt{25}\) which is not a true statement.
We will look at a few different types of equations and then there will be a Quick Check at the end with the different types all put together.
Radical Equations with One Radical
Radical Equations with Two Radicals
Radical Equations with Two Radicals and a Constant Term
Equations with Rational Exponents
A quick note about signs of solutions that tend to confuse students. When we see an equation like \(x^2=25\), we can correctly say that the solutions of that equation are \(x=\pm5\) since squaring a positive or negative \(5\) would give us \(25\). However, the equation \(x=\sqrt{25}\) only gives a solution of \(x=5\) since there is just a positive sign in front of the square root. Some of the extraneous solutions we will get below come into play when you end up with something like \(-5=\sqrt{25}\) which is not a true statement.
We will look at a few different types of equations and then there will be a Quick Check at the end with the different types all put together.
Radical Equations with One Radical
Radical Equations with Two Radicals
Radical Equations with Two Radicals and a Constant Term
Equations with Rational Exponents
Quick Check
Solve equation equation over the real numbers.
- \(\sqrt{2x+14}=\sqrt{x^2+7x}\)
- \(\sqrt{2x+7}+1=\sqrt{x+7}\)
- \(\left(x-7\right)^{\frac{2}{5}}-5=-1\)
- \(\sqrt{x+11}-5=x\)
Quick Check Solutions