rationalequationtransformabletoquadratic

rationale quation transformable to quadratic

1. MATHEMATICS 9 Solving Equations Transformable to Quadratic Equation Including Rational Algebraic Equations Less on 1 : S olvi n g Qua d r a tic E q ua t io n s Th a t Are Not W ri t ten I n S t an d a r d Fo r m Lesson 2 : Solving R ationa l A lgebrai c Equations T r a ns f o r ma b l e T o Qu a d r ati c Eq u ati o ns

2. In solving quadratic equation that is not written in standard form , transform the equation in the standard form ax 2 + bx + c = where a , b , and c are real numbers and a ≠ 0 and then, solve the equation using any method in solving quadratic equation (extracting square roots, factoring, completing the square, or quadratic formula). Les s on 1 : S o lv i ng Quadrat i c E q uations That Are N ot W ri t ten I n S tand a rd Form

3. Example 1: Solve x ( x – 5) = 36. Solution : Transform the equation in standard form. x ( x – 5) = 36 x 2 – 5 x = 36 x 2 – 5 x – 36 = Solve the equation using any method. By factoring x 2 – 5 x – 36 = ( x – 9)( x + 4) = x – 9 = x = 9 x + 4 = x = – 4 The solution set of the equation is {9, – 4} .

4. Example 2: Solve ( x + 5) 2 + ( x – 2) 2 = 37. Solution : Transform the equation in standard form. ( x + 5) 2 + ( x – 2) 2 = 37 x 2 + 10 x + 25 + x 2 – 4 x + 4 = 37 2 x 2 + 6 x + 29 = 37 2 x 2 + 6 x – 8 = x 2 + 3 x – 4 = → 2 x 2 + 6 x + 29 – 37 = Divide all terms by 2. Solve the equation using any method. By factoring x 2 + 3 x – 4 = 0 ( x + 4)( x – 1) = x + 4 = x = – 4 x – 1 = x = 1 The solution set of the equation is {– 4, 1} .

5. Example 3: Solve 2 x 2 – 5 x = x 2 + 14. Solution : Transform the equation in standard form. 2 x 2 – 5 x = x 2 + 14 2 x 2 – x 2 – 5 x – 14 = 0 x 2 – 5 x – 14 = Solve the equation using any method. By Quadratic Formula, identify the values of a , b , and c a = 1, b = -5, c = -14 𝑥 = 2𝑎 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 −(−5) ± (−5) 2 −4(1)(−14) 2 ( 1) = 5 ± 25 + 56 2 = 2 = 5 ± 81 5 ± 9 2 𝑥 = 5+9 = 14 = 𝟕 2 2 𝑥 = 2 5−9 = 2 −4 = −𝟐 The solution set of the equation is {– 2, 7} .

6. Example 4: Solve ( x – 4) 2 = 4. Solution : Transform the equation in standard form. ( x – 4) 2 = 4 x 2 – 8 x + 16 = 4 x 2 – 8 x + 16 – 4 = x 2 – 8 x + 12 = Solve the equation using any method. By factoring x 2 – 8 x + 12 = ( x – 6)( x – 2) = x – 6 = x = 6 x – 2 = x = 2 The solution set of the equation is {6, 2} .

7. Example 5: Solve (3 x + 4) 2 – ( x – 1) 2 = – 5. → 8 x 2 + 26 x + 15 + 5 = Solution : Transform the equation in standard form. (3 x + 4) 2 – ( x – 1) 2 = – 5 9 x 2 + 24 x + 16 – ( x 2 – 2 x + 1) = – 5 9 x 2 + 24 x + 16 – x 2 + 2 x – 1 = – 5 8 x 2 + 26 x + 15 = – 5 8 x 2 + 26 x + 20 = 4 x 2 + 13 x + 10 = Divide all terms by 2. Solve the equation using any method. By Quadratic Formula, identify the values of a , b , and c a = 4, b = 13, c = 10 𝑥 = 2 𝑎 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 −(13) ± (13) 2 −4(4)(10) 2 ( 4 ) = − 1 3 ± 169 − 160 = − 1 3 ± 8 8 = 9 −13 ± 3 8 𝑥 = −13+3 = −10 = − 𝟓 8 8 𝟒 𝑥 = −13−3 = −16 = −𝟐 8 8 𝟒 The solution set of the equation is {– 2, − 𝟓 } .

8. Less on 2 : S olvi n g Ration a l Algeb r ai c E qu a tio n s T ra n sfo r mable T o Quad r ati c E q uati o ns There are rational equations that can be transformed into quadratic equation of the form ax 2 + bx + c = 0 where a, b and c are real numbers, and a ≠ and it can be solved using the different methods in solving quadratic equation .

9. Steps in Solving Rational Equations: Multiply both sides of the equation by the Least Common Multiple (LCM) or Least Common Denominator (LCD). Write the resulting quadratic equation in standard form. Solve the equation using any method in solving quadratic equation. Check whether the obtained values of x satisfies the given equation.

10. Example 1: Solve the rational algebraic equation 6 + 𝑥 4 𝑥−3 = 2 . Solution : 1. Multiply both side of the equation by the LCD, the LCD is 4 x . 4𝑥 6 + 𝑥−3 𝑥 4 = 4 𝑥 ( 2 ) → 4(6) + x ( x – 3) = 8 x 24 + x 2 – 3 x = 8 x 2. Transform the resulting equation in standard form. 24 + x 2 – 3 x = 8 x → x 2 – 3 x – 8 x + 24 = x 2 – 11 x + 24 = 3. Solve the equation using any method. Since the equation is factorable, x 2 – 11 x + 24 = ( x – 3)( x – 8) = x – 3 = x = 3 x – 8 = x = 8 The solution set of the equation is {3, 8} .

11. 1 Example 2: Solve the rational algebraic equation 1 + = 𝑥 𝑥 + 1 12 7 . Solution : 1. Multiply both side of the equation by the LCD, the LCD is 12 x ( x + 1) . 12𝑥(𝑥 + 1) 1 + 1 𝑥 𝑥 + 1 = 12𝑥 𝑥 + 1 7 12 → 12( x + 1) + 12 x = x ( x +1)(7) 12 x + 12 + 12 x = 7 x 2 + 7 x 2. Transform the resulting equation in standard form. 12 x + 12 + 12 x = 7 x 2 + 7 x → = 7 x 2 + 7 x – 12 x – 12 x – 12 = 7 x 2 – 17 x – 12 7 x 2 – 17 x – 12 = 3. Solve the equation using any method. By Quadratic Formula, identify the values of a , b , and c . a = 7, b = – 17, c = – 12 𝑥 = −𝑏± 𝑏 2 −4𝑎𝑐 = −(−17)± (−17) 2 −4(7)(−12) = 17± 289+336 = 17± 625 = 17±25 2𝑎 2(7) 14 14 14 𝑥 = 𝑥 = 1 7 + 2 5 = 4 2 = 𝟑 1 7 − 2 5 = − 8 = − 𝟒 1 4 1 4 1 4 1 4 𝟕 𝟕 The solu t i o n set of t he e qua t i o n i s {3, − 𝟒 } .

12. 8 Example 3: Solve the rational algebraic equation 𝑥 + = 1 + 𝑥 − 2 𝑥 − 2 4𝑥 . Solution : 1. Multiply both side of the equation by the LCD, the LCD is x – 2 . 𝑥 − 2 𝑥 + = 𝑥 − 2 1 + 8 4 𝑥 𝑥 − 2 𝑥 − 2 → x ( x – 2) + 8 = 1( x – 2) + 4 x x 2 – 2 x + 8 = x – 2 + 4 x x 2 – 2 x + 8 = 5 x – 2 2. Transform the resulting equation in standard form. x 2 – 2 x + 8 = 5 x – 2 → x 2 – 2 x – 5 x + 8 + 2 = x 2 – 7 x + 10 = 3. Solve the equation using any method. Since the equation is factorable, x 2 – 7 x + 10 = ( x – 5)( x – 2) = x – 5 = x = 5 x – 2 = x = 2 The solution set of the equation is {5, 2} .

13. Example 4: Solve the rational algebraic equation 𝑥+3 + 1 3 𝑥 − 3 = 4 . Solution : 1. Multiply both side of the equation by the LCD, the LCD is 3( x – 3) . 3(𝑥 − 3) 𝑥+3 + 1 3 𝑥 − 3 = 3(𝑥 − 3)(4) → ( x – 3)( x + 3) + 3(1) = 12( x – 3) x 2 – 9 + 3 = 12 x – 36 2. Transform the resulting equation in standard form. x 2 – 6 = 12 x – 36 → x 2 – 12 x – 6 + 36 = x 2 – 12 x + 30 = 3. Solve the equation using any method. By Quadratic Formula, identify the values of a , b , and c . a = 1, b = – 12, c = 30 𝑥 = −𝑏± 𝑏 2 −4𝑎𝑐 = −(−12)± (−12) 2 −4(1)(30) = 12± 144−120 = 12± 24 = 12±2 6 2𝑎 2(1) 2 2 2 𝑥 = 1 2 + 2 1 2 − 2 2 2 6 = 𝟔 + 𝟔 𝑥 = 6 = 𝟔 − 𝟔 The solution set of the equation is 𝟔 + 𝟔 , 𝟔 − 𝟔 .

14. Example 4: Solve the rational algebraic equation 𝑥 2 3 𝑥 + 2 𝑥 + 1 = . Solution : 1. Multiply both side of the equation by the LCD, the LCD is (3 x + 2)( x + 1) . 𝑥 3 𝑥 + 2 (3𝑥 + 2)(𝑥 + 1) = (3𝑥 + 2)(𝑥 + 1) 2 𝑥 + 1 → x ( x + 1) = 2(3 x + 2) x 2 + x = 6 x + 4 2. Transform the resulting equation in standard form. x 2 + x = 6 x + 4 → x 2 + x – 6 x – 4 = x 2 – 5 x – 4 = 3. Solve the equation using any method. By Quadratic Formula, identify the values of a , b , and c . a = 1, b = – 5, c = – 4 𝑥 = 2 𝑎 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 −(−5) ± (−5) 2 −4(1)(−4) 2 ( 1) = 5 ± 25 + 16 2 = 5 ± 41 2 𝑥 = 𝟓+ 𝟒𝟏 𝟐 𝑥 = 𝟓− 𝟒𝟏 𝟐 The solution set of the equation is 𝟓+ 𝟒𝟏 , 𝟓− 𝟒𝟏 . 𝟐 𝟐