Quadratic Equations PDF

# QUADRATIC EQUATIONS ## Unit-1 In this unit, students will learn how to: - define quadratic equation - solve a quadratic equation in one variable by factorization - solve a quadratic equation in one variable by completing square - derive quadratic formula by using method of completing square - so...

# QUADRATIC EQUATIONS ## Unit-1 In this unit, students will learn how to: - define quadratic equation - solve a quadratic equation in one variable by factorization - solve a quadratic equation in one variable by completing square - derive quadratic formula by using method of completing square - solve a quadratic equation by using quadratic formula - solve the equations of the type ax² + bx + c = 0 by reducing it to the quadratic form - solve the equations of the type ap(x) + b/p(x) = c - solve reciprocal equations of the type a(x² + 1/x² + bx + 1/x) + c = 0 - solve exponential equations involving variables in exponents - solve equations of the type (x + a)(x + b)(x + c)(x + d) = k, where a + b = c + d - solve radical equations of the types - ax + b = cx + d - √x + a + x + b = x + c - √x² + px + m + √x² + px + n = q ## 1.1 Quadratic Equation An equation which contains the square of the unknown (variable) quantity, but no higher power, is called a quadratic equation or an equation of the second degree. A second degree equation in one variable x of the form ax² + bx + c = 0 where a ≠ 0 and a, b, c are real numbers, is called the general or standard form of a quadratic equation. Here a is the co-efficient of x², b is the co-efficient of x and constant term is c. The equations x² - 7x + 6 = 0 and 3x² + 4x = 5 are the examples of quadratic equations. x² - 7x + 6 = 0 is in standard form but 3x² + 4x = 5 is not in standard form. If b = 0 in a quadratic equation ax² + bx + c = 0, then it is called a **pure quadratic equation.** For example x² - 16 = 0 and 4x² = 7 are the pure quadratic equations. **Remember that**: If a = 0 in ax² + bx + c = 0, then it reduces to a linear equation bx + c = 0. **Activity**: Write two pure quadratic equations. ## 1.2 Solution of quadratic equations To find the solution set of a quadratic equation, the following methods are used: - factorization - completing square ### 1.2(i) Solution by factorization: In this method, write the quadratic equation in the standard form as ax² + bx + c = 0 If two numbers r and s can be found for the equation such that r + s = b and rs = ac, then ax² + bx + c can be factorized into two linear factors. **Example 1**: Solve the quadratic equation 3x² - 6x = x + 20 by factorization. **Solution**: 3x² - 6x = x + 20 The standard form of (i) is 3x² - 7x - 20 = 0 Here a = 3, b = -7, c = -20 and ac = 3 * -20 = -60 As -12 + 5 = -7 and -12 * 5 = -60, so the equation (ii) can be written as 3x² - 12x + 5x - 20 = 0 3x(x - 4) + 5(x - 4) = 0 (x - 4)(3x + 5) = 0 Either x - 4 = 0 or 3x + 5 = 0, that is, x = 4 or 3x = -5 ⇒ x = -5/3 Thus, the solution set is { -5/3, 4} **Activity**: Factorize x² -x - 2 = 0 **Example 2**: Solve 5x² = 30x by factorization. **Solution**: 5x² = 30x 5x² - 30x = 0 which is factorized as 5x(x - 6) = 0 Either 5x = 0 or x - 6 = 0 ⇒ x = 0 or x = 6 .. x = 0, 6 are the roots of the given equation. Thus, the solution set is {0, 6}. **Remember that**: Cancelling of x on both sides of 5x² = 30x means the loss of one root i.e., x = 0 ### 1.2(ii) Solution by completing square: To solve a quadratic equation by the method of completing square is illustrated through the following examples. **Example 1**: Solve the equation x² – 3x − 4 = 0 by completing square. **Solution**: x² - 3x - 4 = 0 Shifting constant term -4 to the right, we have x² - 3x = 4 Adding the square of 1/2 x coefficient of x, that is, (-3/2)² on both sides of equation (ii), we get x² - 3x + (-3/2)² = 4 + (-3/2)² or (x - 3/2)² = 4 + 9/4 = 16/4 + 9/4 (x - 3/2)² = 25/4 Taking square root of both sides of the above equation, we have √(x - 3/2 )² = ±√25/4 ⇒ x - 3/2 = ±5/2 x - 3/2 = 5/2 or x - 3/2 = -5/2 Either x = 3/2 + 5/2 = 8/2 = 4 or x = 3/2 - 5/2 = -2/2 = -1 .. 4 and -1 are the roots of the given equation. Thus, the solution set is {-1, 4}. **Example 2**: Solve the equation 2x² – 5x - 3 = 0 by completing square. **Solution**: 2x² - 5x - 3 = 0 Dividing each term by 2, to make coefficient of x² equal to 1. x² - 5/2x - 3/2 = 0 Shifting constant term -3/2 to the right x² - 5/2x = 3/2 Multiply co-efficient of x 1/2 with i.e., 5/2 / 2 = 5/4 Now adding (5/4)² on both sides of the equation (i), we have x² - 5/2x + (-5/4)² = 3/2 + (-5/4)², that is, (x - 5/4)² = 3/2 + 25/16 = 24/16 + 25/16 (x - 5/4)² = 49/16 Taking square root of both sides of the above equation, we have √(x - 5/4)² = ±√49/16 ⇒ x - 5/4 = ±7/4 x - 5/4 = 7/4 or x - 5/4 = -7/4, that is, x = 5/4 + 7/4 = 12/4 = 3 or x = 5/4 - 7/4 = -2/4 = -1/2 .. x = 3, 1/2 are the roots of the given equation. Thus, the solution set is {-1/2, 3}. **Remember that**: For our convenience, we make the co-efficient of x² equal to 1 in the method of completing square. ## 1.3 Quadratic Formula: ### 1.3. (i) Derivation of quadratic formula by using completing square method. The quadratic equation in standard form is ax² + bx + c = 0, a≠0 Dividing each term of the equation by a, we get x² + b/a x + c/a = 0 Shifting constant term to the right, we have x² + b/a x = -c/a Adding (b/2a)² on both sides, we obtain x² + b/a x + (b/2a)² = -c/a + (b/2a)² or (x + b/2a)² = -c/a + b²/4a² Taking square root of both sides, we get √(x + b/2a)² = ±√(b² - 4ac)/4a² or ⇒ x + b/2a = ±√(b² - 4ac)/2a x = ±√(b² - 4ac)/2a - b/2a = -b ±√(b² - 4ac)/2a Thus, x = -b ±√(b² - 4ac)/2a, a≠0 is known as "quadratic formula". **Activity**: Using quadratic formula, write the solution set of x² + x - 2 = 0 ### 1.3 (ii) Use of quadratic formula: The quadratic formula is a useful tool for solving all those equations which can or can not be factorized. The method to solve the quadratic equation by using quadratic formula is illustrated through the following examples. **Example 1**: Solve the quadratic equation 2 + 9x = 5x² by using quadratic formula. **Solution**: 2 + 9x = 5x² The given equation in standard form can be written as 5x² -9x - 2 = 0 Comparing with the standard quadratic equation ax² + bx + c = 0, we observe that a = 5, b = -9, c = -2 Putting the values of a, b and c in quadratic formula x = -b ±√(b² - 4ac)/2a, we have x = -(-9) ±√((-9)² - 4(5)(-2))/2(5) or x = 9 ±√(81 + 40)/10 = 9 ±√121/10 = 9 ±11/10 Either x = 9 + 11/10 = 20/10 = 2 or x = 9 - 11/10 = -2/10 = -1/5, that is, x = 2 or x = -1/ 5 ## 1.4 Equations reducible to quadratic form We now discuss different types of equations, which can be reduced to a quadratic equation by some proper substitution. ### Type (i) The equations of the type ax⁴ + bx² + c = 0 Replacing x² = y in equation ax⁴ + bx² + c = 0, we get a quadratic equation in y. **Example 1**: Solve the equation x⁴ – 13x² + 36 = 0. **Solution**: Let x⁴ - 13x² + 36 = 0 x² = y. Then x⁴ = y² Equation (i) becomes y² - 13y + 36 = 0 which can be factorized as y² - 9y - 4y + 36 = 0 y (y - 9) - 4(y - 9) = 0 (y - 9) (y - 4) = 0 Either y - 9 = 0 or y - 4 = 0, that is, y = 9 or y = 4 Put y = x² x² = 9 or x² = 4 x = ±3 or x = ±2 The solution set is {±2, ±3} ### Type (ii) The equations of the type ap(x) + b/p(x) = c **Example 2**: Solve the equation 2(2x - 1) + 3/(2x - 1) = 5. **Solution**: Given that 2(2x - 1) + 3/(2x - 1) = 5. Let 2x - 1 = y. Then the equation (i) becomes 2y + 3/y = 5 or 2y² + 3 = 5y 2y² - 5y + 3 = 0 Using quadratic formula y = -(-5) ±√((-5)² - 4 * 2 * 3) / 2 * 2 y = 5 ±√(25 - 24)/4 = 5 ±√1/4 = 5 ± 1/4 We have y = 5 + 1/4 = 21/4 or y = 5 - 1/4 = 19/4 When y = 21/4, ⇒ 2x - 1 = 21/4 (. y = 2x - 1) 2x = 21/4 + 1 = 5/4 ⇒ x = 5/8 When y = 19/4, ⇒ 2x - 1 = 19/4 (. y = 2x - 1) 2x = 19/4 + 1 = 23/4 ⇒ x = 23/8 Thus, the solution set is {5/8, 23/8} ### Type (iii) Reciprocal equations of the type: a(x² + 1/x²) + b(x + 1/x) + c = 0 or ax⁴ + bx³ + cx² + bx + a = 0 An equation is said to be a reciprocal equation, if it remains unchanged, when x is replaced by 1/x Replacing x by 1/x in ax⁴ - bx³ + cx² – bx + a = 0, we have a(1/x⁴) - b(1/x³) + c(1/x²) - b(1/x) + a = 0 which is simplified as a - bx + cx² - bx³ + ax⁴ = 0. We get the same equation. Thus ax⁴ – bx³ + cx² − bx + a = 0 is a reciprocal equation. The method for solving reciprocal equation is illustrated through an example. **Example 3**: Solve the equation 2x⁴ - 5x³ – 14x² – 5x + 2 = 0. **Solution**: 2x⁴ - 5x³ - 14x² – 5x + 2 = 0 Dividing each term by x² 2x⁴/x² - 5x³/x² - 14x²/x² - 5x/x² + 2/x² = 0 2x² - 5x -14 - 5/x + 2/x² = 0 2(x² + 1/x²) - 5(x + 1/x) - 14 = 0 Let x + 1/x = y. Then x² + 1/x² = y² - 2 So equation (i) becomes 2(y² - 2) - 5y - 14 = 0 or 2y² - 4 - 5y - 14 = 0 2y² - 5y - 18 = 0 2y² - 9y + 4y - 18 = 0 or y(2y - 9) + 2(2y - 9) = 0 (2y - 9) (y + 2) = 0 Either 2y - 9 = 0 or y + 2 = 0 As y = x + 1/x, so we have 2(x + 1/x) - 9 = 0 2x² + 2 - 9x = 0 2x2 - 9x + 2 = 0 By quadratic formula, we get x= -(- 9) ±√((- 9)² - 4 * 2 * 2) / 2 * 2 x = 9 ±√(81 - 16) / 4 x = 9 ±√ 65 / 4 or or x + 1/x + 2 = 0 x² + 2x + 1 = 0 x = -2 ±√(2)² - 4 * 1 * 1 / 2 * 1 x = -2 ±√4 - 4 / 2 x = - 2 ± 0 / 2 ⇒ x = -1, -1 Thus, the solution set is {-1, -1, (9 - √65)/4, (9 + √65)/4} ### Type (iv) Exponential equations: In exponential equations, variables occur in exponent. The method of solving such equations is illustrated through an example. **Example 4**: Solve the equation 5^{1 + x} + 5^{1 - x} = 26. **Solution**: 5^{1 + x} + 5^{1 - x} = 26 5¹.5^x + 5¹.5^-x = 26 or 5.5^x + 5/5^x = 26 = 0 Let 5^x = y. Then equation (i) becomes 5y + 5/y = 26 = 0 5y² + 5 - 26y = 0 5y² - 25y - y + 5 = 0 5y(y - 5) - 1(y - 5) = 0 (y - 5)(5y - 1) = 0 Either y - 5 = 0 or 5y - 1 = 0, that is, y = 5 or 5y = 1 ⇒ y = 1/5 Put y = 5^x 5^x = 5¹ or 5^x = 5^-1 ⇒ x = 1 or x = -1 .. The solution set {±1}. ### Type (v) The equations of the type: (x + a)(x + b)(x + c)(x + d) = k, where a + b = c + d **Example 5**: Solve the equation (x − 1)(x + 2)(x + 8)(x + 5) = 19. **Solution**: (x − 1)(x + 2)(x + 8)(x + 5) = 19 or [(x - 1)(x + 8)][(x + 2)(x + 5)] – 19 = 0 (x² + 7x-8)(x² + 7x + 10) – 19 = 0 (. -1 + 8 = 2 + 5) Let x² + 7x = y Then eq. (i) becomes (y - 8)(y + 10) – 19 = 0 y² + 2y - 99 = 0 y² + 2y - 80 - 19 = 0 y² + 11y - 9y - 99 = 0 y(y + 11) - 9(y + 11) = 0 (y + 11)(y - 9) = 0 Either y + 11 = 0 or y - 9 = 0 Put y = x² + 7x, so x² + 7x + 11 = 0 Solving by quadratic formula, we have x = -7 ±√(7)² – 4(1)(11) / 2(1) = -7 ± √49 – 44 / 2 = -7 ± √5 / 2 or x² + 7x - 9 = 0 x = -7 ±√(7)² – 4(1)(-9) / 2(1) = -7±√49 + 36 / 2 = -7±√85 / 2 .. The solution set is {-7 ±√5 / 2, -7 ±√85 / 2} ## EXERCISE 1.3 Solve the following equations. 1. 2x⁴ +11x² + 5 = 0 2. 2x⁴ = 9x² - 4 3. 5x^1/2 = 7x^1/4 - 2 4. x^2/3 + 54 = 15x^1/3 5. 3x^-2 + 5 = 8x^-1 6. (2x² + 1) + 2x² + 1 = 4 7. 3/x² + 4/x³ = 4 8. 4x⁺¹ + 4x^-1 = 26 9. (x - a)/(x + a) + (x + a)/(x - a) = 7 10. x⁴ - 2x³ +2x² + 2x + 1 = 0 11. 2x⁴ + x³ - 6x² + x + 2 = 0 12. 4.2^2x+1 - 9.2^x + 1 = 0 13. 3^2x+2 - 12.3^x - 3 = 0 14. 2^x + 64.^2-x - 20 = 0 15. (x + 1)(x + 3)(x - 5)(x - 7) = 192 16. (x -1)(x - 2)(x - 8)(x + 5) + 360 = 0 ## 1.5 Radical equations An equation involving expression under the radical sign is called a **radical equation.** e.g., √x + 3 = x + 1 and √x - 1 = √x - 2 + 1 ### 1.5 (i) Equations of the type: √ax + b = cx + d **Example 1**: Solve the equation √3x + 7 = 2x + 3. **Solution**: √3x + 7 = 2x + 3 Squaring both sides of the equation (i), we get (√3x + 7)² = (2x + 3)² or 3x + 7 = 4x² + 12x + 9 Simplifying the above equation, we have 4x² + 9x + 2 = 0 Applying quadratic formula, x = -9 ±√(9)² - 4*4*2 / 2*4 = -9 ±√81 - 32 / 8 = -9 ±√49 / 8 = -9 ± 7 / 8 x = (-9 + 7) / 8 = -2/8 = -1/4 or x = (-9 - 7) / 8 = -16 / 8 = -2 **Checking**: Putting x = -1/4 in the equation (i), we have √3 (-1/4) + 7 = 2(-1/4) + 3 ⇒√(-3/4 + 7) = -1/2 + 3 ⇒ √25/4 = 5/2 which is true. Putting x = -2 in the equation (i), we get √3(-2) + 7 = 2(-2) + 3 ⇒ √1 = −1 which is not true. On checking, we find that x = -2 does not satisfy the equation (i), so it is an extraneous root. Thus the solution set is {-1/4}. ### 1.5 (ii) Equations of the type √x + a + √x + b = √x + c **Example 2**: Solve the equation √x + 3 + √x + 6 = √x + 11. **Solution**: √x + 3 + √x + 6 = √x + 11 Squaring both sides of the equation (i), we have x + 3 + x + 6 + 2(√x + 3)(√x + 6) = x + 11 or 2√x² + 9x + 18 = - x + 2 Squaring both sides of the equation (ii), we get 4(x² + 9x + 18) = x² -4x + 4 or 3x² + 40x + 68 = 0 Applying quadratic formula, we get x = - 40 ±√(40)² - 4 * 3 * 68 / 2 * 3 = - 40 ±√1600 - 816 / 6 We have x = (- 40 + √784)/ 6 = -40 ±28 / 6 x = (-40 + 28) / 6 = -12 / 6 = -2 or x = (-40 - 28) / 6 = -68 / 6 = -34 / 3 **Checking**: Putting x = -2 in the equation (i), we have √ -2 + 3 + √x - 2 + 6 = √ -2 + 11 √-34 + 9 + √ -34 + 18 = √ -34 + 11 √1 + √16 = √33 √3 x √(-1) + √16 x √(- 1) = √1 x √(-1) As -34/3 is extraneous root, so the solution set is {-2}. **Note**: Extraneous root is introduced either by squaring the given equation or clearing it of fractions. ### 1.5(iii) Equations of the type: √x² + px + m + √x² + px + n = q **Example 3**: Solve the equation √x² – 3x + 36 - √x² – 3x + 9 = 3. **Solution**: √x² - 3x + 36 - √x² – 3x + 9 = 3 Let x² - 3x = y Then √y + 36 - √y + 9 = 3 Squaring both sides, we get y + 36 + y + 9 – 2(√y + 36)(√y + 9) = 9 2y + 45 - 2√(y + 36)(y + 9) = 9 -2y² + 45y + 324 = - 2y - 36 √y² + 45y + 324 = y + 18 Again squaring both sides, we get y² + 45y + 324 = y² + 36y + 324 9y = 0 ⇒ y = 0 or -2y² + 45y + 324 = -2(y + 18) As x² - 3x = y, so x² – 3x = 0 ⇒ x(x - 3) = 0 Either x = 0 or x - 3 = 0 x = 3 .. x = 0, 3 are the roots of the equation. Thus, the solution set is {0, 3}. ## EXERCISE 1.4 Solve the following equations. 1. 2x + 5 = √7x + 16 2. √x + 3 = 3x - 1 3. 4x^1/3 + 13x^1/6 + 14 = 3 4. √3x + 100 - x = 4 5. √x + 5 + √x + 21 = x + 60 6. √x + 1 + √x - 2 = √x + 6 7. √11 - x - √6 - x = √27 - x 8. √4a + x - √a - x = √a 9. √x² + x + 1 - √x² + x - 1 = 1 10. √x² + 3x + 8 + √x² + 3x + 2 = 3 11. √x² + 3x + 9 + √x² + 3x + 4 = 5 ## MISCELLANEOUS EXERCISE - 1 ### 1. Multiple Choice Questions Four possible answers are given for the following questions. Tick (√) the correct answer. (i) Standard form of quadratic equation is - bx + c = 0, b ≠ 0 - ax² + bx + c = 0, a ≠ 0 - ax² = bx, a ≠ 0 - ax² = 0, a ≠ 0 (ii) The number of terms in a standard quadratic equation ax² + bx + c = 0 is - 1 - 2 - 3 - 4 (iii) The number of methods to solve a quadratic equation is - 1 - 2 - 3 - 4 (iv) The quadratic formula is - x = -b ±√(b² - 4ac) / 2a - x = -b ±√(b² + 4ac) / 2a - x = b ±√(b² + 4ac) / 2a - x = b ±√(b² - 4ac) / 2a (v) Two linear factors of x² - 15x + 56 are - (x -7) and (x + 8) - (x + 7) and (x - 8) - (x - 7) and (x - 8) - (x + 7) and (x + 8) (vi) An equation, which remains unchanged when x is replaced by 1/x is called a/an - Exponential equation - Reciprocal equation - Radical equation - None of these (vii) An equation of the type 3x + 3^2-x + 6 = 0 is a/an - Exponential equation - Radical equation - Reciprocal equation - None of these (viii) The solution set of equation 4x² - 16 = 0 is - {±4} - {4} - {±2} - ±2 (ix) An equation of the form 2x⁴ - 3x³ + 7x² - 3x + 2 = 0 is called a/an - Reciprocal equation - Radical equation - Exponential equation - None of these ### 2. Write short answers of the following questions. (i) Solve x² + 2x - 2 = 0 (ii) Solve by factorization 5x² = 15x (iii) Write in standard form 1/(x+4) + 1/(x - 4) = 3 (iv) Write the names of the methods for solving a quadratic equation. (v) Solve (2x - 1)² / 9 = 2 / 4. (vi) Solve √3x + 18 = x. ### 3. Fill in the blanks (i) The standard form of the quadratic equation is _______. (ii) The number of methods to solve a quadratic equation are ______. (iii) The name of the method to derive a quadratic formula is ______. (iv) The solution of the equation ax² + bx + c = 0, a ≠ 0 is ______. (v) The solution set of 25x²- 1 = 0 is ______. (vi) An equation of the form 2^2x - 3.2^x + 5 = 0 is called a/an ______ equation. (vii) The solution set of the equation x² - 9 = 0 is ______. (viii) A root of an equation, which do not satisfy the equation is called a/an ______ root. (ix) An equation of the type x⁴ + x³ + x² + x +1 = 0 called a/an ______ equation. (x) An equation involving impression of the variable under ______ is called radical equation. ## SUMMARY An equation which contains the square of the unknown (variable) quantity, but no higher power, is called a quadratic equation or an equation of the second degree. A second degree equation in one variable x, ax² + bx + c = 0 where a ≠ 0 and a, b, c are real numbers, is called the general or standard form of a quadratic equation. An equation is said to be a reciprocal equation, if it remains unchanged, when x is