Solving Rational Equations and Non-linear inequalities

Solving Rational Equations and Non-linear inequalities:

Solving Rational Equations and Non-linear inequalities

Rational Equations:

Rational Equations Simply put, rational equations are equations that contain polynomials in fractions. To solve, multiply each term by the common denominator, which eliminates the fractions, then simplify and solve. You may have to factor, use the Rational Root Theorem, use square roots, use the Quadratic Formula, etc. Each equation will be different. Be sure to check for extraneous solutions. Use the values which would make the fraction undefined to help you. Those are values that x cannot be. If you get one as a solution, it’s not a solution.

Example 1:

Example 1 Keep in the back of your mind that x cannot equal 8. Multiply by the common denominator. Simplify. Solve.

Slide 4:

Use the Quadratic Formula.

Example 2:

Example 2 Find the common denominator. Factor down the left side, and you will get (x-2)(x-4). This is your common denominator. Multiply every term by the common denominator. Simplify. Solve. Keep in the back of your mind that x cannot equal 4 or 2.

We will pause for a brief mental break. :

We will pause for a brief mental break.

Solving linear inequalities:

Solving linear inequalities They are similar to solving polynomial equations, but with some differences. Inequalities typically have a range of possible answers, so we have to find the possible ranges as our solution. Find the roots of the polynomial. (Again, you may have to factor, use Rational Root Theorem, use Quadratic Formula, etc.) Depending upon the inequality symbol, these may or may not be included in the solution set. Once you have found the roots, place them on a number line, and test values in each range to see if they are true. Write the solutions using interval notation.

Example 1:

Example 1 Solve. Get everythin g on one side . Solve. Factor the 2 out.

Slide 10:

Put these on a number line. 2 4 You now have three ranges to check: less than 2, between 2 and 4, and greater than 4. Pick a value for each range and plug it into the inequality. (You can use the original or the factored.) If it comes out true, then that range is included in the solution. If it is false, the range is not included. I’m going to use 0, 3, and 5; it does not matter what you pick.

Slide 11:

False, so this range is not a possible solution. True, so this range is a solution. False, so this range is not a possible solution. The only range that worked was the section in between 2 and 4. Since the inequality was , you use the brackets, because the numbers are part of the possible solution. Solution set: [2, 4]

Example 2:

Example 2 This one requires more work. We can’t simply factor or use the Quadratic Formula, so we have to use the Rational Root Theorem. Solve. Rule of Signs: Positive Roots: There is 1 sign change, so 1 positive root. Negative Roots: There is 1 sign change in f(-x), so 1 negative root.

Slide 13:

Use synthetic division to find a root. I’ll try 1 first. 2 9 2 -3 9 -48 -3 Didn’t work. Next random choice: -3

Slide 14:

2 1 -6 -15 -3 0 45 Victory! -3 is our negative root. Since the leading power was 3, that means our new leading power is 2, so I can use the quadratic formula to find the other roots. A = 2, B = 1, C = -15 My roots are -3 and 2.5. Since -3 has multiplicity 2, we only write it once.

Slide 15:

Put these on a number line. -3 2.5 You now have three ranges to check: less than 2, between 2 and 4, and greater than 4. Pick a value for each range and plug it into the inequality. I’m going to use -4, 0, and 3; it does not matter what you pick. I’m going to use synthetic division to find the value; you can just plug in if you wish. Doesn’t matter.

Slide 16:

2 -1 -8 -8 4 -13 32 -13 is not ≥ 0. That range does not work. 2 7 0 12 0 -45 0 -45 is not ≥ 0. That range does not work. Since we know that -3 works, and the inequality symbol is ≥, -3 is a possible solution. You will write it by itself in a set. Now, we must test the final range, using 3.

Slide 17:

2 13 6 27 39 36 81 36 is greater than or equal to 0, so this range works as well. Since the -3 worked as a solution, but neither range around it did, we write it by itself. The range after 2.5 worked, so we put that into brackets. Since the symbol was ≥, it’s brackets for those two, but a parentheses for the infinity symbol.