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Quadratic Equations yield two (2) answers since the highest exponent is 2. It is possible that the two numbers are the same, or they may be different. They may or may not be integers or rational or even Real. There is a method for predicting the "nature" of the answers which is controlled by the "Discriminant", a specific part of the Quadratic Formula. For now, know that it is even possible that the solutions are "Imaginary/Complex Numbers".
There are three basic methods for solving Quadratic Equations:
Factoring Make sure the equation is in "standard form" which means all terms on one side of = and other side 0 and check to see "if" the equation is factorable. This would be the simplest method for solving if, indeed, the equation is easily factorable. Example: Find the roots of x2+3x-10 = 0 The "c" term (-10) makes us think of 5 and 2, and so does the "b" term (3). Try the factors (x+5) and (x-2) to see if these are correct by multiplying together using FOIL or "distributive property". So... and...(x+5)(x-2) = 0 This line is telling us that the two quantities have a product of ZERO, a very special answer only arrived at when one of the original values =0. The factors "lead" to the solutions {-5, 2}. Verify that these are correct by substituting into the original equation, ie. The same result would occur when substituting the value 2 into the original eqauation. Suggestion: A complete review of factoring would be a good idea. You should be familiar with all the different types: Common Factor, Difference of 2 Perfect Square Terms, Square Trinomial, and General Trinomial. Completing the Square In order to perform this method, you must be familiar with what a "Square Trinomial" is and looks like. The a, b, and c terms must be related in just the right way, ie. x2 + 14x + 49. Both 14 and 49 make us think "7" and the trinomial factors into identical factors of (x+7)(x+7). Also, x2 -22x + 121 makes us think "11" and (x-11)(x-11). x2 - 22x + 50 does not have the same affect. One big "clue" is that 50 is not itself a perfect square number. This clue is exactly what leads us to the procedure for this method. Example: Find the roots of We really want "121" which comes from (-22/2)2 not "50" so add 71 to BOTH sides and get... and (x-11)2 = 71 Now take the square root (sqrt) of both sides and get... and x = 11 + - sqrt 71. If the answers were going to be "pretty" the equation would have been factorable in the first place. You may use this method on ANY Quadratic Equation, however it is best saved for those with the "a" term = 1 and the "b" term being even or else the math gets a little tricky (fractions and arithmetic). Hence, the need for another method. Quadratic Formula Any quadratic equation of the form ax2+bx+c=0 can be solved using the quadratic formula. In all cases the two solutions of x are given by the formula:
Example: Find the roots of After the equation is put into "standard form" we get a = 1, b = -4, and c = 3. These terms are then substituted into the quadratic formula (perhaps by careully using a caluculator) to arrive at the two solutions. Here is what the first substitution looks like. (4 + (sqrt(16-12))/2 (4 + (sqrt(4))/2 (4 + 2)/2 (6)/2 = 3 So...3 is one of the solutions. Discriminant One little part of the Quadratic Formula is the predictor for the "nature" of the solutions(roots) for quadratic equations. The value of (b2-4ac) IS the predictor. Since the square root of this value must be found, we have three cases which could evolve.
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