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Published byEvangeline Watson Modified over 6 years ago

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Table of Contents First note this equation has "quadratic form" since the degree of one of the variable terms is twice that of the other. When this occurs, begin by making the substitution u = x smaller exponent. Here, Equation Of Quadratic Form: Solving Algebraically Example: Solve 4x 4 – 35x 2 – 9 = 0. u = x 2. This means u 2 = x 4. Substitute these into 4x 4 – 35x 2 – 9 = 0 to get: 4u 2 – 35u – 9 = 0. Next, solve this equation by factoring or using the quadratic formula.

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Table of Contents Last, solve each of the resulting equations. In this case they are solved by taking the square root of both sides. Equation Of Quadratic Form: Solving Algebraically Slide 2 Next replace each u with x 2 so becomes Try to solve 2x 2/3 – 11x 1/3 + 12 = 0. The solutions are x = 64 and x =

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Table of Contents Equation Of Quadratic Form: Solving Algebraically

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