How to Solve Algebra II Equations

Equations that arise in Algebra II include zeros of quadratic formulae, inequalities, absolute value equations and systems of linear equations. There are others, but these provide a solid foundation for whatever other types of equations one may encounter.

Things You'll Need

• Calculator

Instructions

Quadratic equations

1. Learn the quadratic formula by heart. It comes up too often and is too useful not to.

   The solution of quadratic equations of the form ax^2 + bx + c = 0 is [-b +/- sqrt(b^2 - 4ac)] / [2a] and is called the quadratic formula.
   sqrt denotes square root and +/- signifies plus or minus. Therefore, two solutions are possible.

2. Factor the equation if you can.

   Search for easy factoring to avoid having to use the lengthy quadratic formula. Making numerous calculations increases the chance of error.
   If the quadratic equation is easily factorable--such as x^2 - 9 = 0 or x^2 + 2x + 1 = 0--the quadratic formula need not be solved.
   For example, these two equations factor into (x-3)(x+3) = 0 and (x+1)(x+1) = 0.

3. Solve for the individual parts once the equation is factored.
   For example, in (x-3)(x+3) = 0, one of the two products must equal 0. Therefore, x-3 = 0 may be a solution and x+3 = 0 may be a solution.

4. Solve for the possible solutions.
   Taking the above example to its conclusion, x-3 = 0 means x = 3, and x+3 = 0 means x = -3. So x equals either 3 or -3.

5. If factoring is not clear, use the quadratic formula to solve for x.

Inequalities

1. Use the usual algebraic manipulations to isolate x on one side to solve an inequality for unknown x.
   Keep in mind that when you divide by a negative number, the inequality reverses. For example, -3 < 2, but 3 > -2.
   So as an algebraic example, suppose you're solving (3x-2) < 1. Isolate x by adding 2 to both sides, then by dividing through by 3 to get x < (1+2) / 3. Therefore x < 1.

2. Split the inequality as follows when the algebraic manipulations are not enough, for example, for a higher order:
   Suppose instead of the inequality in Step 1, you're given (x-2)^2 < 1. To take the root of both sides as your first step in isolating x, you must account for the possibility that x-2 is negative. If it is, then it still has to be greater than -1 because no negative number less than -1 can square to be less than 1.
   So we can split the inequality to get -1 < x-2 < 1.

3. Proceed with the usual methods of algebraic manipulation to isolate x further, just applying the operations to all three elements of the equation:
   Adding through by 2: 1 < x < 3.

Absolute value equations

1. Try to isolate x, but as with inequalities, split the equation into two when both positive and negative values are possible.

   For example, for |x-3| = 9, account must be taken of x-3 being negative and x-3 being positive. If the former, then x-3 = -9, and the solution is x = -6. If x-3 is positive, just solve x-3 = 9.

2. Break up the equation, more than once if necessary, if there is more than one set of absolute value brackets.

   For example, for ||x-3| + |x+4|| = 11, it must be considered whether |x-3| + |x+4| is positive or negative (fortunately it is always positive, so we can just remove its absolute value brackets). Then consideration is made whether x-3 is positive or not and whether x+4 is positive or not.

3. Make a line diagram of critical values if that would be helpful.

   The critical values for ||x-3| + |x+4|| = 11 are at x=3 and x=-4, since that's when the signs of the arguments of the absolute value brackets change.

4. Fill out the line with the signs of the arguments of the absolute value brackets.

   Below -4, both x-3 and x+4 are negative. So the equation becomes (3-x) + (-x-4) = 11, or x = -6.

   Between -4 and 3, x-3 is negative and x+4 is positive. So the equation becomes (3-x) + (x+4) = 11, for which x has no solution.

   Above 3, the equation merely needs all absolute value brackets removed, since everything is positive. So the equation becomes (x-3) + (x+4) = 11, or x = 5.

Systems of equations

1. Eliminate one variable to get the value of the other variable by adding or subtracting scalar products of the equations from each other.

   For example, for the system

   x + y = 1
   2x - y = 2

   add the two equations to cancel out the y's, to get

   (x+y) + (2x-y) = 1 + 2

   or

   3x = 3.

   So x = 1.

2. Plug the value of the known variable back into one of the equations, to solve for the other variable's value.

   Continuing from the example in Step 1:

   (1) + y = 1

   Therefore, y = 0.

3. Solve for three variables in three equations by performing more steps, adding scalar products of equations, to achieve the desired isolation of each variable one-by-one.

References

• Algebra II: Quadratic equations
• Algebra II: Solving Equations and Inequalities