
Solution - DrDelMath
... The quadratic equation obtained by squaring both sides of the original equation need not be equivalent to the original equation. Therefore both of these possible solution must be tested. I asked you to not perform those tests because of the complexity. 28. Solve the equation x2 = 5x – 6 x2 = 5x – 6 ...
... The quadratic equation obtained by squaring both sides of the original equation need not be equivalent to the original equation. Therefore both of these possible solution must be tested. I asked you to not perform those tests because of the complexity. 28. Solve the equation x2 = 5x – 6 x2 = 5x – 6 ...
Practice Test 2 – Topics
... 1. Factoring: Only in special cases two terms, three terms, four terms 2 real roots/solutions only set factors = 0 independently 2. Quadratic formula Can be used on any quadratic equation 0, 1, 2 real or complex roots/solutions 3. Completing the square Can be used on any quadratic equatio ...
... 1. Factoring: Only in special cases two terms, three terms, four terms 2 real roots/solutions only set factors = 0 independently 2. Quadratic formula Can be used on any quadratic equation 0, 1, 2 real or complex roots/solutions 3. Completing the square Can be used on any quadratic equatio ...
File
... intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and ...
... intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and ...
MATH 117 The Development of Complex Numbers
... Unfortunately, the algebra most often does not work out so nicely so as to give the desired solution. In fact, how did we know in advance to choose 2 + −1 so that ...
... Unfortunately, the algebra most often does not work out so nicely so as to give the desired solution. In fact, how did we know in advance to choose 2 + −1 so that ...