## Presentation on theme: "QUADRATIC EQUATIONS."— Presentation transcript:

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.

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 (This makes it quadratic) Similarly equation with degree one is a linear equation. Eg: 3X=9 General form: ax²+bx+c=0

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 a

Example 2 Take a simple example of a ball dropped from a height
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

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

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.

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

Example 2 : Solve the equation (m + 8)² –5 = 31 Solution : (m + 8)² –5 = 31 (m + 8)² = 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

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 = √( ) solving h = √(144) taking the roots h = ±12 ans h = +12 or h = -12 Are the roots

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

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

Example 3 : Solve the quadratic equation 4k (3k – 1) = 5
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

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

Derivation

Problems(Formula)

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

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

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.

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

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 = 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 = (–4)2 –4 (4) (1) Δ = 16 – 16 ∴ Δ = 0 Roots are real and equal

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

Sum and product of the roots

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

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

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

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

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

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