PP Section 2.2 - Other Types of Equations In this section...

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Unformatted text preview: Other Types of Equations In this section we review how to solve several types of equations. Example: Absolute Value Solve the equation 3 x + 1 = 7 3x + 1 = 7 3x + 1 = 7 ⇒ 3 x + 1 = −7 2 3x = 6 ⇒ ⇒x= 8 3 x = −8 − 3 Example: Absolute Value Solve the equation 4 x − 2 = 3x − 4 Solving the equation above is equivalent to finding the x­ coordinates of the points of intersection of the two curves in the plot. 4 x − 2 = 3x − 4 x − 2, x − 2 ≥ 0 OR x ≥ 2 x−2 = − x + 2, x − 2 < 0 OR x < 2 4( x − 2), x ≥ 2 ⇒ 3x − 4 = 4 x − 2 = 4(− x + 2), x < 2 4( x − 2), x ≥ 2 3x − 4 = 4 x − 2 = 4(− x + 2), x < 2 ⇒ 3 x − 4 = 4 x − 8, x ≥ 2 ⇒ x = 4 12 ⇒ 3 x − 4 = −4 x + 8, x < 2 ⇒ 7 x = 12 ⇒ x = 7 Checking answers: 4 x − 2 = 3x − 4 x = 4 : 4 4 − 2 = 3(4) − 4 ⇒ 8 = 8 12 12 88 12 x = : 4 − 2 = 3 − 4 ⇒ = 7 7 77 7 Example: Roots Solve the equation ( x 5 − 1) 3 + 8 = 0 ( x − 1) + 8 = 0 ⇒ ( x − 1) = −8 ⇒ x − 1 = −2 5 3 5 3 5 x = 1 − 2 = −1 ⇒ x = −1 5 Example: Roots Solve the equation ( x 2 − 1) 3 + 8 = 0 ( x − 1) + 8 = 0 ⇒ ( x − 1) = −8 ⇒ x − 1 = −2 2 3 2 3 2 x = 1 − 2 = −1 ⇒ NO real solution 2 Example: Quadratic Type Solve the equation x 4 − 2 x 2 − 14 = 0 Completing the square: x 4 − 2 x 2 − 14 = ( x 4 − 2 x 2 + 1) − 15 = ( x 2 − 1) 2 − 15 = 0 ⇒ ( x − 1) = 15 ⇒ x − 1 = ± 15 ⇒ x = 1 ± 15 2 2 2 2 1 − 15 < 0 ⇒ x = 1 + 15 ⇒ x = ± 1 + 15 2 Example: Quadratic Type Solve the equation x 4 − 2 x 2 − 14 = 0 Using the quadratic formula: 2 ± 4 − 4(−14) 2 ± 60 x= = = 1 ± 15 2 2 2 x ≥ 0 ⇒ x = 1 + 15 ⇒ x = ± 1 + 15 2 2 Example: Radicals Solve the equation 2 + 10 − x = − x From looking at the equation we know: x ≤ 10 and − x ≥ 0 ⇒ x ≤ 0 2 + 10 − x = − x ⇒ 10 − x = − x − 2 ⇒ 10 − x = (− x − 2) = x + 4 x + 4 2 2 ⇒ x 2 + 5 x − 6 = 0 ⇒ ( x + 6)( x − 1) = 0 ( x + 6)( x − 1) = 0 ⇒ x = −6 O R x = 1 x ≤ 10 and − x ≥ 0 ⇒ x ≤ 0 ⇒ x = −6 Check answer: 2 + 10 − x = − x 2 + 10 − (−6) = 2 + 16 = 2 + 4 = 6 = −(−6) Note: x = 1 is an extraneous solution, a solution that appears in the algebraic process, but is NOT a solution of the original problem. Example: Fractional Exponents Solve the equation x + 3 x 3 − 28 = 0 4 3 2 x + 3 x − 28 = ( x ) + 3x − 28 = u + 3u − 28 4 3 2 3 2 3 2 2 3 2 u + 3u − 28 = (u + 7)(u − 4) = 0 2 ⇒ u = −7 OR u = 4 u = x ⇒ u ≥ 0 ⇒ x 3 = 4 2 3 2 x = 4 ⇒ x = (±2) = ±8 2 3 3 ...
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This note was uploaded on 01/22/2011 for the course MATH 2 taught by Professor Gardner during the Fall '08 term at Irvine Valley College.

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