A level Maths: Pure Y1 - Algebraic expression and Quadratics

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= 0, where a, b, c are real number and fundi function a +0. ● To solve quadratic equations, a given equation must be rewritten in the form or kept in the form of αx²+bx+c = 0. . ax² +bx+c Must be factorised Quadratic formula: x = -b ± √√√b²-4ac 2a COMPLETING THE SQUARE: • x² + bx+c = ( x + b₂² ) ²³ - ( b )² + c 2 Example x²+6x+2 (x+6) ² (€) ²+2 2 (x+3)²-9+2 (x+3)²-7 WAYS TO SOLUE: factorisation ↳ Completing the square ↳ quadratic formula FUNCTIONS • A function can be seen as a machine that takes in an input, converts this value Mathema- tically and gives an output. • Input = denoted with 'x' • Output = represented as f(x) or g(x) For a given function, the set of possible inputs is called domain and the set of possible outputs is called the range Function x f(x) 1x² + x ● 3 QUADRATIC GRAPHS 0 f(x) = ax² + bxctc y = ax² + bx + ( ↳a is positive the shape of the parabola would be Lais negative the shape of the parabola would be H (B) Intersection of the y-axis ACSO: Minimum value of a function is the place where the graph has a verter at its lowest point. Maximum value of a function is the place where the graph has a vertex at its highest point. •A and B roots 62 Turning point are the intersection of the x-axis at these points y=o 3²+x = 11 f(x= (x+b)²-(1)² + c La stationary point ↳ can be found by completing the square :.T.P - (-2,- () ² + c) THE DISCRIMINANT The expression b²-4ac is called the discriminant. 4 • The value obtained using the discriminant expression on a quadratic function will indicate how many roots a function f(x) has. given part of the quadratic formula - b ± √(b) ² - 4 Cac) 20 ● For the quadratic function f(x) = ax² + bx+c. * if the discrimant b²- 4ac >0 then the function f(x) has 2 distinct real roots. * if the discrimant b²- 4ac =0, then the function f(x) has I repeated real root. * If the discriminant b² (ac <0, then the function f(x) has no real roots. EXAMPLE x = - find the range of U for which x² + 2x +h=0 has 2 distinct real roots. x² + 2x + 1 = 0 a = 1, b = 2 7 b²-4ac>0 for 2 distinct real roots 2²- 4 x 1 x 4 >0 2²-4K>0 C = K -4K>-4 K21 HIDDEN QUADRATICS Example Solve x4- 13x² + 36 = 0 f(x) = x², so let y = x ² and substitute into the equation: y) 13y + 36 = as x ² = (x²)² = y² 10 Then, solve by factorising (y- 9) (y-4)= 0 y-9=0 ⇒ y = 9 y - 4 = 0 ⇒ y = 4 (3 marks) x= ± √3 x= ±2