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• Title page, A Presentation by Dr. Brian Burns.
• The Quadratic Formula uses the coefficients a, b and c of the standard form of a Quadratic Equation to calculate the solutions. The Quadratic Formula is obtained by completing the square for a general Quadratic Equation in Standard Form. The Discriminant is defined as the part you take the square root of, it is labeled D and is equal to b squared minus the product of 4, a and c. The solutions are given by dividing the negative of b plus or minus the square root of the discriminate by two times a. If the discriminate is zero then plus or minus zero is the same, so you get one real solution (the negative of b divided by two times a). If the discriminate is greater than zero, you get two different real solutions. If the discriminant is less than zero, you are taking the square root of a negative number and hence have no real solutions.
• Here we solve a quadratic equation, already in standard form, with coefficients 6, 11 and 3. With b equal to 11, b squared is 121, with a equal to 6 and c equal to 3, the product of 4, a and c gives 72. The discriminant is then 121 minus 72 which is equal to 49, which is 7 squared. This example has a positive discriminant, whose square root is an integer, the value 7. The negative of b is negative 11. If we subtract 7 then divide by two a, we get negative 18 over 12 or negative three halves as one solution. If we add 7 to negative 11 and divide by two a, we get negative 4 over 12 or negative one third. These are the two elements of the solution set.
• Here we solve a quadratic equation, by first putting into standard form, with coefficients 2, negative 6 and 3. With b equal to negative 6, b squared is 36, with a equal to 2 and c equal to 3, the product of 4, a and c gives 24. The discriminant is then 36 minus 24 which is equal to 12, which is 4 times 3. This example has a positive discriminant, whose square root simplifies to 2 times the square root of 3. The negative of b is 6. Negative b minus the square root of the discriminant has a common factor of two, which cancels out when dividing by 4, the value of two times a. The solution set is then 3 plus or minus the square root of 3, all divided by 2.
• Here we solve a quadratic equation, by first putting into standard form, with coefficients 4, 8 and 4. With b equal to 8, b squared is 96, with a equal to 4 and c equal to 4, the product of 4, a and c gives 96. The discriminant is then zero. If we had removed the common factor of 4 first, we would get coefficients 1,2 and 1, which also give a zero discriminant. The calculations are simpler if we use a equal to 1, b equal to 2 and c equal to 1. Either way we get a complete square with solution negative b divided by two times a, giving negative 1 as the only element of the solution set.
• Here we solve a quadratic equation, by first putting into standard form, with coefficients 5, 6 and 3. With b equal to 6, b squared is 36, with a equal to 5 and c equal to 3, the product of 4, a and c gives 60. The discriminant is then 36 minus 60 which is equal to minus 24, which less than zero. There are hence no real solutions and the solution set consists of the empty set.
• A summary of procedures. After putting into standard form and removing any common factors, you find the values of the coefficients a, b and c. You then use these to calculate the discriminant, b squared minus 4 times a, times c. The value of the discriminant determines if you have one real solution, two real solutions or no real solutions. If solutions exist you use the quadratic formula to find the solutions. In general, once in standard form, you should factor if you can see an easy way to factor. If you cannot easily find a factorization, you should solve by completing the square or using the quadratic formula.