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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 4x4 – 35x2 – 9 = 0.

u = x2.

This means u2 = x4. Substitute these into 4x4 – 35x2 – 9 = 0 to get: 4u2 – 35u – 9 = 0.

Next, solve this equation by factoring or using the quadratic formula.

41

,9 uandu

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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 x2 so 41

,9 uandu

becomes .41

,9 22 xandx

ixandx21

,3

Try to solve 2x2/3 – 11x1/3 + 12 = 0.

The solutions are x = 64 and x = .827

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Equation Of Quadratic Form: Solving Algebraically