Tiêu đề 01 Quadratic Equation(1).pdf ✅

QUADRATIC EQUATION 1 CHAPTER CONTENTS • Basic Concept • Value of a Quadratic Polynomial • Zeros of a Quadratic Polynomial • Quadratic Equation & Its Roots • Solving a Quadratic Equation By Factorisation • Solving a quadratic Equation By Completing the square • Solving a Quadratic Equation By Using Quadratic Formula • Nature or Character of the Roots of a Quadratic Equation • Sum and Product of The Roots • To Construct A Quadratic Equation Whose Roots Are Given • Equation Reducible to Quadratic Equations • Problems on Quadratic Equations ➢ BASIC CONCEPT  Every algebraic polynomial of second degree is called a quadratic polynomial. For Example : (i) 3x2 + 5x + 7 (ii) 8x2 – 6x (iii) 5x2 – 7 (iv) 2 x 2 + 6x – 3  The general form of quadratic polynomial is ax2 + bx + c; where a, b, c are real numbers, a  0 and x is variable. For a particular quadratic polynomial the values of a, b, c are constant and for this reason a, b and c are also called real constants. For example, in quadratic polynomial 3x2 – 5x + 8; 3, – 5 and 8 are constant where as x is variable. ➢ VALUE OF A QUADRATIC POLYNOMIAL The value of a quadratic polynomial ax2 + bx + c (i) at x =  is a() 2 + b() + c = a2 + b + c (ii) at x =  is a 2 + b + c (iii) at x = 5 is a (5)2 + b(5) + c = 25a2 + 5a + c In the same way : (i) Value of 5x2 – 3x + 4 at x = 2 is = 5(2)2 – 3(2) + 4 = 20 – 6 + 4 = 18 (ii) Value of x2 – 8x – 15 at x = – 1 is = (–1)2 – 8 (–1) – 15 = 1 + 8 – 15 = – 6 (iii) Value of 7x2 – 4 at x = 3 2 is = 7 2 3 2       – 4 = 7 × 9 4 – 4 = 9 28 − 36 = 9 8
➢ ZEROS OF A QUADRATIC POLYNOMIAL The value of the polynomial x2 – 7x + 10 at : (i) x = 1 is (1)2 – 7 × 1 + 10 = 1 – 7 + 10 = 4 (ii) x = 2 is (2)2 – 7 × 2 + 10 = 4 – 14 + 10 = 0 (iii) x = 3 is (3)2 – 7 × 3 + 10 = 9 – 21 + 10 = – 2 (iv) x = 5 is (5)2 – 7 × 5 + 10 = 25 – 35 + 10 = 0 It is observed here that for x = 2 and x = 5; the value of polynomial x2 –7x + 10 is zero. These two values of x are called zeros of the polynomial. Thus, if for x = , where  is a real number, the value of given quadratic polynomial is zero; the real number  is called zero of the quadratic polynomial. ❖ EXAMPLES ❖ Ex.1 Show that : (i) x = 3 is a zero of quadratic polynomial x 2 – 2x – 3. (ii) x = – 2 is a zero of quadratic polynomial 3x2 + 7x + 2. (iii) x = 4 is not a zero of quadratic polynomial 2x2 – 7x – 5. Sol.(i) The value of x2 – 2x – 3 at x = 3 is (3)2 – 2 × 3 – 3 = 9 – 6 – 3 = 0  x = 3 is a zero of quadratic polynomial x2 – 2x – 3. (ii) The value of 3x2 + 7x + 2 at x = – 2 is 3(–2)2 + 7 (–2) + 2 = 12 – 14 + 2 = 0  x = – 2 is a zero of quadratic polynomial 3x2 + 7x + 2 (iii) The value of 2x2 – 7x – 5 at x = 4 is 2(4)2 – 7(4) – 5 = 32 – 28 – 5 = – 1  0  x = 4 is not a zero of quadratic polynomial 2x2 – 7x – 5. Ex.2 Find the value of m, if x = 2 is a zero of quadratic polynomial 3x2 – mx + 4. Sol. Since, x = 2 is a zero of 3x2 – mx + 4  3(2)2 – m × 2 + 4 = 0  12 – 2m + 4 = 0, i.e., m = 8. ➢ QUADRATIC EQUATION & IT’s ROOTS Since, ax2 + bx + c, a  0 is a quadratic polynomial, ax2 + bx + c = 0, a  0 is called a quadratic equation. (i) –x 2 – 7x + 2= 0 is a quadratic equation, as –x 2 – 7x + 2 is a quadratic polynomial. (ii) 5x2 – 7x = 0 is a quadratic equation. (iii) 5x2 + 2 = 0 is a quadratic equation, but (iv) –7x + 2 = 0 is not a quadratic equation. ❖ EXAMPLES ❖ Ex.3 Which of the following are quadratic equations, give reason : (i) x 2 – 8x + 6 = 0 (ii) 3x2 – 4 = 0 (iii) 2x + x 5 = x2 (iv) x 2 + 2 x 2 = 3 Sol. (i) Since, x2 – 8x + 6 is a quadratic polynomial  x 2 – 8x + 6 = 0 is a quadratic equation. (ii) 3x2 – 4 = 0 is a quadratic equation. (iii) 2x + x 5 = x3  2x2 + 5 = x3  x 3 – 2x2 – 5 = 0; which is cubic and not a quadratic equation. (iv) x 2 + 2 x 2 = 3  x 4 + 2 = 2x2  x 4 – 2x2 + 2 = 0; which is biquadratic and not a quadratic equation. Ex.4 In each of the following, determine whether the given values are solutions (roots) of the equation or not : (i) 3x2 – 2x – 1 = 0; x = 1 (ii) x 2 + 6x + 5 = 0; x = – 1, x = – 5 (iii) x 2 + 2 x – 4 = 0; x = 2 , x = – 2 2 Sol. (i)  Value of 3x2 – 2x – 1 at x = 1 is
3(1)2 – 2(1) – 1 = 3 – 2 – 1 = 0 = RHS  x = 1 is a solution of the given equation. (ii) For x = – 1, L.H.S. = (–1)2 + 6 (–1) + 5 = 1 – 6 + 5 = 0 = R.H.S.  x = – 1 is a solution of the given equation For x = – 5, L.H.S. = (–5)2 + 6(–5) + 5 = 25 – 30 + 5 = 0 = R.H.S.  x = – 5 is a solution of the given equation. (iii) For x = 2 , L.H.S. = x2 + 2 x – 4 = ( 2 ) 2 + 2 ( 2 ) – 4 = 2 + 2 – 4 = 0 = R.H.S.  x = 2 is a solution of the given equation For x = – 2 2 , L.H.S. = (–2 2 ) 2 + 2 × – 2 2 – 4 = 4 × 2 – 2 × 2 – 4 = 0 R.H.S.  x = – 2 2 is a solution of the given equation. SOLVING A QUADRATIC EQUATION BY FACTORISATION ➢ Since, 3x2 – 5x + 2 is a quadratic polynomial; 3x2 – 5x + 2 = 0 is a quadratic equation. Also, 3x2 – 5x + 2 = 3x2 – 3x – 2x + 2 [Factorising] = 3x (x – 1) – 2(x – 1) = (x – 1) (3x – 2) In the same way : 3x2 – 5x + 2 = 0  3x2 – 3x – 2x + 2 = 0 [Factorising L.H.S.]  (x – 1) (3x – 2) = 0 i.e., x – 1 = 0 or 3x – 2 = 0  x = 1 or x = 3 2 ; which is the solution of given quadratic equation. In order to solve the given Quadratic Equation: 1. Clear the fractions and brackets, if given. 2. By transfering each term to the left hand side; express the given equation as ; ax2 + bx + c = 0 or a + bx + cx2 = 0 3. Factorise left hand side of the equation obtained (the right hand side being zero). 4. By putting each factor equal to zero; solve it. ❖ EXAMPLES ❖ Ex.5 Solve : (i) x 2 + 3x – 18 = 0 (ii) (x – 4) (5x + 2) = 0 (iii) 2x2 + ax – a 2 = 0; where ‘a’ is a real number. Sol. (i) x 2 + 3x – 18 = 0  x 2 + 6x – 3x – 18 = 0  x(x + 6) – 3(x + 6) = 0 i.e., (x + 6) (x – 3) = 0  x + 6 = 0 or x – 3 = 0  x = – 6 or x = 3  Roots of the given equation are : – 6 and 3 (ii) (x – 4) (5x + 2) = 0 x – 4 = 0 or 5x + 2 = 0  x = 4 or x = – 5 2 (iii) 2x2 + ax – a 2 = 0  2x2 + 2ax – ax – a 2 = 0  2x(x + a) – a(x + a) = 0 i.e., (x + a) (2x – a) = 0  x + a = 0 or 2x – a = 0  x = – a or x = 2 a Ex.6 Solve the following quadratic equations : (i) x 2 + 5x = 0 (ii) x 2 = 3x (iii) x2 = 4 Sol. (i) x 2 + 5x = 0  x(x + 5) = 0  x = 0 or x + 5 = 0  x = 0 or x = – 5 (ii) x 2 = 3x
 x 2 – 3x = 0  x(x – 3) = 0  x = 0 or x = 3 (iii) x 2 = 4  x = ± 2 Ex.7 Solve the following quadratic equations : (i) 7x2 = 8 – 10x (ii) 3(x2 – 4) = 5x (iii) x(x + 1) + (x + 2) (x + 3) = 42 Sol. (i) 7x2 = 8 – 10x  7x2 + 10x – 8 = 0  7x2 + 14x – 4x – 8 = 0  7x(x + 2) – 4(x + 2) = 0  (x + 2) (7x – 4) = 0  x + 2 = 0 or 7x – 4 = 0  x = – 2 or x = 7 4 (ii) 3(x2 – 4) = 5x  3x2 – 5x – 12 = 0  3x2 – 9x + 4x – 12 = 0  3x(x – 3) + 4(x – 3) = 0  (x – 3) (3x + 4) = 0  x – 3 = 0 or 3x + 4 = 0  x = 3 or x = – 3 4 (iii) x(x + 1) + (x + 2) (x + 3) = 42  x 2 + x + x2 + 3x + 2x + 6 – 42 = 0  2x2 + 6x – 36 = 0  x 2 + 3x – 18 = 0  x 2 + 6x – 3x – 18 = 0  x(x + 6) – 3(x + 6) = 0  (x + 6) (x – 3) = 0  x = – 6 or x = 3 Ex.8 Solve for x : 12 abx2 – (9a2 – 8b2 ) x – 6ab = 0 Sol. Given equation is : 12abx2 – 9a2x + 8b2x – 6ab = 0  3ax(4bx – 3a) + 2b(4bx – 3a) = 0  (4bx – 3a) (3ax + 2b) = 0  4bx – 3a = 0 or 3ax + 2b = 0  x = 4b 3a or x = – 3a 2b SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE ➢ Every quadratic equation can be converted in the form : (x + a)2 – b 2 = 0 or (x – a)2 – b 2 = 0. Steps : 1. Bring, if required, all the term of the quadratic equation to the left hand side. 2. Express the terms containing x as x2 + 2xy or x 2 – 2xy. 3. Add and subtract y2 to get x2 + 2xy + y2 – y 2 or x2 – 2xy + y2 – y 2 ; which gives (x + y)2 – y 2 or (x – y)2 – y 2 . Thus, (i) x 2 + 8x = 0  x 2 + 2x × 4 = 0  x 2 + 2x × 4 + 42 – 4 2 = 0  (x + 4)2 – 16 = 0 (ii) x 2 – 8x = 0  x 2 – 2 × x × 4 = 0  x 2 – 2 × x × 4 + 42 – 4 2 = 0  (x – 4)2 – 16 = 0 ❖ EXAMPLES ❖ Ex.9 Find the roots of the following quadratic equations (if they exist) by the method of completing the square. (i) 2x2 – 7x + 3 = 0 (ii) 4x2 + 4 3 x + 3 = 0 (iii) 2x2 + x + 4 = 0 Sol. (i) 2x2 – 7x + 3 = 0  x 2 – 2 7 x + 2 3 = 0 [Dividing each term by 2]  x 2 – 2 × x × 4 7 + 2 3 = 0