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QUADRATIC EQUATIONS

QUADRATIC EQUATIONS. Objectives. Identify a quadratic equation. Distinguish between a pure and an adfected quadratic equation. Solve simple problems on pure and adfected quadratic equations. Identify the standard form of a quadratic equation.

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QUADRATIC EQUATIONS

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  1. QUADRATIC EQUATIONS

  2. Objectives • Identify a quadratic equation. • Distinguish between a pure and an adfected quadratic equation. • Solve simple problems on pure and adfected quadratic equations. • Identify the standard form of a quadratic equation. • Solve the quadratic equations by factorization. • Solve the quadratic equations by using formula. • Find the value of the discriminant and know the nature of the roots. • Frame the quadratic equation for the given roots. • Solve the quadratic equation graphically.

  3. Definition and explanation • The name Quadratic comes from “quad(in Latin)” meaning square, because the variable gets squared(X²). • Also an equation of second order or degree 2 is quadratic equation. Eg: X²–3X+2 (This makes it quadratic) Similarly equation with degree one is a linear equation. Eg: 3X=9 General form: ax²+bx+c=0

  4. a Example 1 Consider a square of side ‘a’ units and its area 25 square units. Area of the square = (side)² 25 = a² or a²= 25 ∴ a =±5 i.e. a=+5 or a=-5 A Quadratic equation has only two roots. a

  5. Example 2 Take a simple example of a ball dropped from a height. According to physics any particle moving with uniform acceleration the equation is given as ½at²+ut-s=0 This is a quadratic equation having the form ax²+bx+c=0

  6. Types of quadratic equations(Pure and Adfected) Quadratic equation involving a variable only in second degree is a “Pure Quadratic Equation’’. From the general form when b=0 we get ax² + c = 0, where a and c are real numbers and a ≠ 0 is a pure quadratic equation. Eg: (1) x² – 9 = 0 (2) 2a² – 18 = 0

  7. Quadratic equation involving a variable in second degree as well as in first degree is an “Adfected Quadratic Equation”. ax2 + bx + c = 0 is the standard form of a quadratic equation where a, b and c are variables and a ≠ 0. Eg: (1) x2 + 3x – 10 = 0 (2) 3a2 – a – 2 = 0 Thus a easy way to differentiate between the 2 types is when b = 0 and b ≠ 0.

  8. Problems (Pure) Example 1 : Solve the equation 3X²-27=0 Solution : 3X² – 27 = 0 ∴ 3X² = 27 get 27 to RHS X² = 27/3 divide it by 3 X² = 9 result is 9 ∴X² = 9 Find the square root X = ±3 Ans X=+3 or X= – 3 Are the roots of the equation

  9. Example 2 : Solve the equation (m + 8)² –5 = 31 Solution : (m + 8)² –5 = 31 (m + 8)² = 31 + 5 get 5 to RHS (m + 8)² = 36 add it to 31 ∴ (m + 8) = √36 square root of 36 (m + 8) = ±6 roots ∴ m = –8 ± 6 simplify m = –8 + 6 or m = –8 – 6 m = -2 or m= -14 ans Are the roots of equation

  10. Example 3 : If l2 = r2 + h2. Solve for h and find the value of ‘h’ if l = 15 and r =9 Solution : l2 = r2 + h2 r² + h² = l² h² = l² – r² solving for h h = √(l² - r²) h = √(15² - 9²) substitute the values h = √(255 - 81) solving h = √(144) taking the roots h = ±12 ans h = +12 or h = -12 Are the roots

  11. Problems(Adfected) Example 1 : Solve the quadratic equation a2 – 3a + 2 = 0 Solution : a2 – 3a + 2 = 0 a² – 2a – 1a + 2 = 0 Resolve the expression a(a – 2) –1(a – 2) = 0 Factorize (a – 2) (a – 1) = 0 Take common factor a – 2= 0 or a – 1 = 0 Equate each factor to 0 i.e. a = 2 or a = 1 Are the roots

  12. Example 2 : Solve the quadratic equation 2x² – 3x + 1 = 0 Solution : 2x² – 3x + 1 = 0 2x² – 2x – 1x + 1= 0 Resolve the expression 2x (x – 1) –1(x – 1)=0 Factorize (x – 1) (2x – 1) = 0 Take common factor (x – 1) = 0 or (2x – 1) = 0 Equate each factor to 0 x = 1 or x = ½ Are the roots

  13. Example 3 : Solve the quadratic equation 4k (3k – 1) = 5. Solution : 4k (3k – 1) = 5 12k² – 4k – 5 = 0 12k² – 10k + 6k – 5 = 0 2k (6k – 5) + 1(6k – 5) = 0 (6k – 5) (2k + 1) = 0 (6k – 5) = 0 or (2k + 1) = 0 k = 5/6 or k = -½

  14. Solving the equation Consider the equation x² + 3x + 1 = 0 It cannot be factorised by splitting the middle term. How do you solve such an equation ? It can be solved by using Formula. Derivation is as follows

  15. Derivation

  16. Problems(Formula)

  17. Problems(Factorization) Example 1 : Solve the equation (x + 6) (x + 2) = x Solution :(x + 6) (x + 2) = x x² + 6x + 2x + 12 = x Resolve the expression x² + 8x + 12 – x = 0 Factorize x² + 7x + 12 = 0 x² + 4x + 3x + 12 = 0 x(x + 4) + 3 (x + 4) = 0 Take common factor (x + 4) (x + 3) = 0 (x + 4) = 0 or (x + 3) = 0 Equate each factor to 0 x = -4 or x = -3 Are the roots

  18. Problems based on Quadratic Equation Example 1 : If the square of a number is added to 3 times the number, the sum is 28. Find the number. Solution : Let the number be = x Square of the number = x² 3 times the number = 3x Square of a number + 3 times the number = 28 x² + 3x = 28 x²+ 3x – 28 = 0 x²+ 7x – 4x – 28 = 0 x(x + 7) –4 (x + 7) = 0 (x + 7) (x – 4) = 0 x + 7 = 0 or x – 4 = 0 x = –7 or x = 4 ∴ The required number is 4 or –7

  19. Nature of the roots and Discriminant After solving the equations we get two roots and there can be three nature of roots: (1)Equal Eg: X = 2 and X = 2 (2)Distinct Eg: X = 1 and X = 2 Eg: X = 1 and X = -2 (3)Imaginary Eg: X = 1 +√(-2) and X = 1- √(-2) Thus for all the problems the roots can be classified into these 3 categories.

  20. It is clear that, 1) Nature of the roots of quadratic equation depends upon the value of (b2 – 4ac) 2) The Expression (b2 – 4ac) is denoted by Δ (delta) which determines the nature of the roots. 3) In the equation ax2 + bx + c = 0 the expression (b2 – 4ac) is called the discriminant. Discriminant (b2 – 4ac) Nature of the roots Δ = 0 Roots are real and equal Δ > 0 (Positive) Roots are real and distinct Δ < 0 (negative) Roots are imaginary

  21. Example 1 : Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0. Consider the equation 2x2 – 5x – 1 = 0 This is in form of ax2 + bx + c = 0 The co-efficient are a = 2, b = –5, c = –1 Δ = b2 – 4ac Δ = (–5)2 –4(2) (–1) Δ = 25 + 8 Δ = 33 ∴ Δ > 0 Roots are real and distinct Example 2 : Determine the nature of the roots of the equation 4x2 – 4x + 1 = 0 Consider the equation 4x2 – 4x + 1 = 0 This is in the form of ax2 + bx + c = 0 The co-efficient are a = 4, b = –4, c = 1 Δ = b2 – 4ac Δ = (–4)2 –4 (4) (1) Δ = 16 – 16 ∴ Δ = 0 Roots are real and equal

  22. Example 3 : For what values of ‘m’ roots of the equation x² + mx + 4 = 0 are (i) equal (ii) distinct Consider the equation x² + mx + 4 = 0 This is in the form ax² + bx + c = 0 the co-efficients are a = 1, b = m, c = 4 Δ = b² – 4ac Δ = m² – 4(1) (4) Δ = m² – 16 (1) If roots are equal Δ = 0 m² – 16 = 0 m² = 16 m = √16 ∴ m = ±4 (2) If roots are distinct Δ > 0 m² – 16 > 0 m² > 16 m > √16 m > ±4

  23. Sum and product of the roots

  24. Formation of quadratic equation If ‘m’ and ‘n’ are the roots then the Standard form of the equation is x² – (Sum of the roots) x + Product of the roots = 0 x² – (m+ n) x + mn = 0 Let ‘m’ and ‘n’ are the roots of the equation ∴ x = ‘m’ or x = ‘n’ i.e., x – m = 0, x – n = 0 (x – m) (x – n) = 0 ∴ x² – mx – nx + mn = 0 x² – (m + n) x+ mn = 0

  25. Example 1 : Form the quadratic equation whose roots are 2 and 3 Let ‘m’ and ‘n’ are the roots ∴m = 2, n = 3 Sum of the roots = m + n = 2 + 3 ∴ m + n = 5 Product of the roots = mn = (2) (3) ∴ mn = 6 Standard form x² – (m+ n) x+ mn = 0 x² – (5)x + (6) = 0

  26. Example 2 : Form the quadratic equation whose roots are 3 + 2√5 and 3 – 2√5 Let ‘m’ and ‘n’ are the roots ∴ m = 3 + 2√5 and n = 3 – 2√5 Sum of the roots = m + n = 3 + 2√5 + 3 – 2√5 ∴ m + n = 6 Product of the roots = mn = (3 + 2√5 ) (3 – 2√5 ) = (3)² –( 2√5 )² = 9 – 20 ∴ mn = – 11 x² – (m + n) x + mn = 0 ∴ x² – 6x – 11 = 0

  27. Example 3 : If ‘m’ and ‘n’ are the roots of equation x² – 3x + 4 = 0 form the equation whose roots are m² and n². Solution: Consider the equation x² – 3x + 4 = 0 The coefficients are a = 1, b = –3, c = 4 Let ‘m’ and ‘n’ are the roots Sum of the roots = m + n = − b/a = − (−3)/1 ∴ m + n = 3 ii) Product of the roots = mn = c/a =4/1 ∴ mn = 4 If the roots are ‘m²’ and ‘n²’ Sum of the roots m² + n² = (m + n)² – 2mn = (3)² – 2(4) = 9 – 8 ∴ m² + n² = 1 Product of the roots m²n² = (mn)² = 4² ∴ m²n² = 16 x² – (m² + n²) x + m²n² = 0 ∴ x² – (1)x + (16) = 0 ∴ x² – x + 16 = 0

  28. Graphical method of solving a Quadratic Equation Graphical method of solving is another way to solve a quadratic equation The graph of a quadratic polynomial is a curve called ‘parabola’ Eg: X² - 3X – 10 Roots are X = -2 and X = 5

  29. THANKYOU

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