Solving Quadratic Equations By Keith Rachels And Asef Haider What is a Quadratic Equation? A quadratic function has the form y= ax^2 + bx + c where “a” does not equal to 0. The quadratic form shown above is written in standard form. In the following slides, there are two other forms of the quadratic function. Vertex Form • The vertex form of a quadratic function is shown as : y= a(x-h)^2 + k - The vertex is (h,k) and the axis of symmetry is x=h Intercept Form *The quadratic formula written in intercept form looks like this: y= a(x-p)(x-q) • In this form, the x-intercepts are p and q. Solving Quadratic Functions in Standard Form • Solve x2 + 5x + 6 = 0. • This equation is in standard form. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero. • So the first thing I have to do is factor: • x2 + 5x + 6 = (x + 2)(x + 3) • Set this equal to zero: • (x + 2)(x + 3) = 0 x + 2 = 0 or x + 3 = 0 x = –2 or x = – 3 • The solution to x2 + 5x + 6 = 0 is x = –3, –2 F.O.I.L • • FOIL means first, outside, inside, last. That's not too hard to remember if you say it in your head a few times. You use FOIL to multiply the terms inside the parenthesis in a specific order: first, outside, inside, last. Here's how to solve (4x + 6)(x + 2): First - multiply the first term in each set of parenthesis: 4x * x = 4x^2 • Outside - multiply the two terms on the outside: 4x * 2 = 8x • Inside - multiply both of the inside terms: 6 * x = 6x • Last - multiply the last term in each set of parenthesis: 6 * 2 = 12 • Now just add everything together to get 4x^2 + 14x + 12. This method only works easily with two binomials. To multiply something complicated like (4x + 6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the answer onto the remaining binomial. • Example using FOIL • • • • Example: Multiply the following: (2x-5)(x-4) Solution: Just follow the letters in FOIL: First: 2x*x=2x^2. Outside: 2x*(-4)=-8x. Inside: -5*x=-5x. Last: (-5)*(-4)=20. • Sum it all up and you get: (2x^2-13x+20). The Quadratic Formula The quadratic formula is an equation that helps you find the roots of an equation you can’t normally factor. Before you apply an equation to the quadratic formula, you must write the equation in standard form, ax2 + bx + c = 0 The Quadratic Formula • The quadratic formula is the equation X = -b ± b2 – 4ac 2a If b2 - 4ac > 0, then the equation has two real solutions If b2 - 4ac = 0, then the equation has one real solution If b2 - 4ac < 0, then the equation has two imaginary solutions • Let’s find the values of x in the equation x2 + 4x + 29 We see that a = 1 b = 4 and c = 29 x = -(4)± (4) 2 – 4(1)(29) 2(1) x = -4 ± 16 – 116 x = -4 ± 10i 2 2 x = -2 + 5i x = -4 ± i 100 x = -2 – 5i 2 Completing the Square • Completing the square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. • To complete the square of x2 + bx, you have to add (b ÷ 2) 2 • The equation looks like this x2 + bx + (b÷2) 2 = (x + (b÷2)) 2 • Let’s find the value of c that makes the expression a perfect square trinomial and then write the equation as the square of the binomial! :D • x2 – 12x + c x2 – 12x +( ) 2 x2 – 12x + (-6) 2 x2 – 12x + 36 (x - 6)(x - 6) (x – 6) 2 Now let’s solve an equation by completing the square! :D x2 + 20x + 104 = 0 x2 + 20x + ( ) 2 = -104 x2 + 20x + (10) 2 = -104 + (10) 2 x2 + 20x + 100 = -4 x + 10 = ± 2i (x + 10) 2 = -4 x = -10 ± 2i x = -10 + 2i (x + 10) 2 = -4 x = -10 – 2i Sum of Two Cubes • The sum of two cubes is represented by the equation (a3 + b3 ) = (a + b)(a2 – ab + b2 ) • Let’s factor the polynomial x3 + 27 x3 + 27 = x3 + 33 = (x + 3)(x2 – 3x + 9) Difference of Two Cubes • The difference of two cubes is represented by the equation (a – b)(a2 + ab + b2 ) • It looks similar to the sum of two cubes but the positive and negative signs a swapped between the b in the first set of parenthesis and the ab in the second set of parenthesis • Sum (a + b)(a2 – ab + b2 ) • Difference (a – b)(a2 + ab + b2 ) • Let’s try to find the difference of two cubes in the equation 16u5 – 250u2 Factor common 16u5 – 250u2 = 2u2(8u3 – 125) monomial = 2u2((2u)3 – 53) (a3 – b3) = 2u2(2u – 5)(4u2 + 10u + 25) (a - b) (a2 + ab + b2)

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