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Published byDamon Cross Modified over 4 years ago

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Other Types of Equations

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Solving a Polynomial Equation by Factoring 1.Move all terms to one side and obtain zero on the other side. 2.Factor. 3. Apply the zero product principle, setting each factor equal to zero. 4. Solve the equations in step 3. 5.Check the solutions in the original equation.

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Text Example Solve by factoring: 3x 4 = 27x 2. Step 1 Move all terms to one side and obtain zero on the other side. Subtract 27x 2 from both sides 3x 4 x 2 27x 2 27x 2 3x 4 27x 2 Step 2 Factor. 3x 4 27x 2 3x 2 (x 2 - 9) 0

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Solution cont. Solve by factoring: 3x 4 = 27x 2. Steps 3 and 4 Set each factor equal to zero and solve each resulting equation. 3x 2 = 0orx 2 - 9 = 0 x 2 = 0x 2 = 9 x = 0x = 9 x = 0x = 3 Steps 5 check your solution

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Example Solve: Answer:

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Radical Equations A radical equation is an equation in which the variable occurs in a square root, cube root or higher root.

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Example Solve: Answer: Solution: Isolate the radical by moving the other terms to the one side Square both sides to remove the radical Move all terms to one side Factor CHECK EACH “ANSWER”!!!! Only one works!!!!

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Solving Radical Equations of the Form x m/n = k Assume that m and n are positive integers, m/n is in lowest terms, and k is a real number. 1.Isolate the expression with the rational exponent. 2.Raise both sides of the equation to the n/m power.

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Solving Radical Equations of the Form x m/n = k cont. If m is even:If m is odd: x m/n = k x m/n = k (x m/n ) n/m = ±k(x m/n ) n/m = k n/m x = ±k n/m x = k n/m It is incorrect to insert the ± when the numerator of the exponent is odd. An odd index has only one root. 3. Check all proposed solutions in the original equation to find out if they are actual solutions or extraneous solutions.

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Text Example Solve: x 2/3 - 3/4 = -1/2. Isolate x 2/3 by adding 3/4 to both sides of the equation: x 2/3 = 1/4. Raise both sides to the 3/2 power: (x 2/3 ) 3/2 = ±(1/4) 3/2. x = ±1/8.

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Some equations that are not quadratic can be written as quadratic equations using an appropriate substitution. Here are some examples. 5t 2 + 11t + 2 = 0t = x 1/3 5x 2/3 + 11x 1/3 + 2 = 0 or 5(x 1/3 ) 2 + 11x 1/3 + 2 = 0 t 2 – 8t – 9 = 0t = x 2 x 4 – 8x 2 – 9 = 0 or (x 2 ) 2 – 8x 2 – 9 = 0 New EquationSubstitutionGiven Equation Equations That Are Quadratic in Form

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Rewriting an Absolute Value Equation without Absolute Value Bars If c is a positive real number and X represents any algebraic expression, then |X| = c is equivalent to X = c or X = -c.

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Example Solve: Answer: 3x-1=4 and 3x-1=-4 solve, 3x=53x=-3 x=5/3 x=-1

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Other Types of Equations

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