Exercise 20 Page 266

Solving a radical equation usually involves three main steps.

1. Isolate the radical on one side of the equation.
2. Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
3. Solve the resulting equation. Remember to check your results!

Now we can analyze the given radical equation. sqrt(3-8x^2)=2x Notice that in this equation there is an isolated radical with index equal to 4 on the left-hand side. Then, we will raise each side of the equation to the power of 4.

We obtained a polynomial equation. To solve it, we will define another variable. If we let z=x^2, we can rewrite the last equation in terms of the z-variable. 16x^4+8x^2-3=0 ⇔ 16z^2+8z-3=0 Note that the above equation in terms of z is a quadratic equation. There are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve the equation. Let's determine a, b, and c. 16z^2+8z-3=0 ⇔ 16z^2+ 8z+( - 3)=0 We see above that a= 16, b= 8, and c= - 3. Let's substitute these values into the Quadratic Formula to solve the equation.

Now we can calculate the first root using the positive sign and the second root using the negative sign.

We found that the solutions for 16z^2+8z-3=0 are z= 14 and z=- 34. This means that x^2= 14 and x^2=- 34. Since x^2 is non-negative, the equation x^2=- 34 has no solutions. Let's solve x^2= 14.

Checking the Solutions

Next, we will check the solutions by substituting 12 and - 12 for x into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution. Let's first check the 12.

Because our substitution produced a true statement, we know that our answer, x= 12, is correct. Let's check now - 12.

Because our substitution produced a false statement, we know that our answer, x= 12, is not correct.