Special Products 1. $ (x + y)(x - y) = x^2 - y^2 $ 2. $ (x + y)^2 = x^2 + 2xy + y^2 $ 3. $ (x - y)^2 = x^2 - 2xy + y^2 $ 4. $ (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 $ 5. $ (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 $ 6. $ (x + a)(x + b) = x^2 + (a + b)x + ab $ 7. $ (ax + by)(cx + dy) = acx^2 + (ad + bc)xy + bdy^2 $ Factoring Numbers have factors.And expressions (like x2+4x+3) also have factors. Factoring (called Factorising in the UK) is the process of finding the factors.Factoring: Finding what to multiply together to get an expression. It is like splitting an expression into a multiplication of simpler expressions. Example: factor 2y+6 Both 2y and 6 have a common factor of 2: 2y is 2 × y 6 is 2 × 3 So you can factor the whole expression into: 2y+6 = 2(y+3) So 2y+6 has been factored into 2 and y+3 Factoring is also the opposite of Expanding..expand vs factor Common Factor In the previous example we saw that 2y and 6 had a common factor of 2 But to do the job properly make sure you have the highest common factor, including any variables Example: factor 3y2+12y Firstly, 3 and 12 have a common factor of 3. So you could have: 3y2+12y = 3(y2+4y) But we can do better! 3y2 and 12y also share the variable y. Together that makes 3y: 3y2 is 3y × y 12y is 3y × 4 So you can factor the whole expression into: 3y2+12y = 3y(y+4y) Check: 3y(y+4) = 3y × y + 3y × 4 = 3y2+12y Solving Quadratic Equation by Completing the Square The quadratic in the previous sections last example, (x – 2)2 – 12, can be multiplied out and simplified to be x2 – 4x – 8. But we would not have been able to solve the equation with the quadratic formatted this way because it doesnt factor and it isnt ready for square-rooting. The only reason we could solve it before was because theyd already put all the x stuff inside a square, so we could square-root both sides. So how do you go from a regular quadratic like x2 – 4x – 8 to one that is ready to be square-rooted? We would have to complete the square. I have a lesson on solving quadratics by completing the square, which explains the steps and gives examples of this process. It also shows how the Quadratic Formula is generated by this process. So Ill just do just one example of the process in this lesson. If you need further instruction, read the lesson at the above hyperlink. Use completing the square to solve x2 – 4x – 8 = 0. As noted above, this quadratic does not factor, so I cant solve the equation by factoring. And they havent given me the quadratic in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that form, and then solve. It works like this: First, I put the loose number on the other side of the equation: x2 – 4x – 8 = 0 x2 – 4x = 8 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), giving me –2. Then I square this value to get +4, and add this squared value to both sides of the equation: x2 – 4x + 4 = 8 + 4 x2 – 4x + 4 = 12 This process creates a quadratic that is a perfect square, and factoring gives me.(x – 2)2 = 12 Now I can square-root both sides of the equation, simplify, and solve: (x – 2)2 = 12 x = 2 ± 2sqrt(3) Then the solution is x = 2 ± 2sqrt(3) Solving Quadratic Equation by extracting square roots Solve x2 – 4 = 0. Previously, Id solved this by factoring the difference of squares, and solving each factor; the solution was x = ± 2. However— I can also try isolating the squared variable term, putting the number over on the other side, like this: x2 – 4 = 0 x2 = 4 I know that, when solving an equation, I can do whatever I like to that equation as long as I do the same thing to both sides of the equation. On the left-hand side of this particular equation, I have an x2, and I need a plain x. To turn an x2 into an x, I can take the square root of each side of the equation: sqrt(x) = ± sqrt(4) x = ± 2 Then the solution is x = ± 2 Solve x2 – 50 = 0. This quadratic has a squared part and a number part. Ill start by adding the numerical term to the other side of the equaion (so the squared part is by itself), and then Ill square-root both sides. Ill need to remember to simplify the square root: x2 – 50 = 0 x2 = 50 x = ± 5sqrt(2) Then the solution is x = ± 5 sqrt(2) Solve (x – 5)2 – 100 = 0. This quadratic has a squared part and a number part. Ill start by adding the number to the other side of the equation, so the squared binomial is by itself. Then Ill square-root both sides, remembering to simplify my results: (x – 5)2 – 100 = 0 (x – 5)2 = 100 sqrt[ (x - 5)^2 ] = ± sqrt(100) x – 5 = ±10 x = 5 ± 10 x = 5 – 10 or x = 5 + 10 x = –5 or x = 15 The solution is x = –5, 15 Solving Quadratic Equations by Quadratic Formula Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the square. At some point, he noticed that he was always doing the exact same steps in the exact same order for every equation. Taking advantage of the one of the great powers and benefits of algebra (namely, the ability to deal with abstractions, rather than having to muck about with the numbers every single time), he made a formula out of what hed been doing: The Quadratic Formula: For ax2 + bx + c = 0, the value of x is given by: x = [-b+/-sqrt(b^2-4ac)]/2a The nice thing about the Quadratic Formula is that the Quadratic Formula always works. There are some quadratics (most of them, actually) that you cant solve by factoring. But the Quadratic Formula will always spit out an answer, whether the quadratic was factorable or not. I have a lesson on the Quadratic Formula, which gives examples and shows the connection between the discriminant (the stuff inside the square root), the number and type of solutions of the quadratic equation, and the graph of the related parabola. So Ill just do one example here. If you need further instruction, study the lesson at the above hyperlink. Use the Quadratic Formula to solve x2 – 4x – 8 = 0. Looking at the coefficients, I see that a = 1, b = –4, and c = –8. Ill plug them into the Formula, and simplify. I should get the same answer as before: x = 2 ± 2sqrt(3) Then the solution is x = 2 ± 2sqrt(3) Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the square. At some point, he noticed that he was always doing the exact same steps in the exact same order for every equation. Taking advantage of the one of the great powers and benefits of algebra (namely, the ability to deal with abstractions, rather than having to muck about with the numbers every single time), he made a formula out of what hed been doing: The Quadratic Formula: For ax2 + bx + c = 0, the value of x is given by: x = [-b+/-sqrt(b^2-4ac)]/2a The nice thing about the Quadratic Formula is that the Quadratic Formula always works. There are some quadratics (most of them, actually) that you cant solve by factoring. But the Quadratic Formula will always spit out an answer, whether the quadratic was factorable or not. I have a lesson on the Quadratic Formula, which gives examples and shows the connection between the discriminant (the stuff inside the square root), the number and type of solutions of the quadratic equation, and the graph of the related parabola. So Ill just do one example here. If you need further instruction, study the lesson at the above hyperlink. Lets try that last problem from the previous section again, but this time well use the Quadratic Formula: Use the Quadratic Formula to solve x2 – 4x – 8 = 0. Looking at the coefficients, I see that a = 1, b = –4, and c = –8. Ill plug them into the Formula, and simplify. I should get the same answer as before: x = 2 ± 2sqrt(3) Then the solution is x = 2 ± 2sqrt(3) Advisory: When using the Formula, make sure you are careful not to omit the ± sign, and be careful with the fraction line (dont draw it as being only under the square root; its under the initial –b part, too). And, though many of your quadratics will start with x2 so a = 1, dont forget that the denominator of the Formula is 2a, not just 2; that is, when the leading term is something like 5x2, you will need to remember to put the a = 5 value in the denominator. Take the time to be careful, because, as long as you do your work neatly, the Quadratic Formula will give you the right answer every time. Nature of the Roots of a Quadratic Equation Discriminant = b² -4ac Case I. When a, b, c are real numbers, a 0: If = b² -4 a c = 0, then roots are equal (and real). If = b² -4 a c > 0, then roots are real and unequal. If = b² -4 a c < 0, then roots are complex. It is easy to see that roots are a pair of complex conjugates. Case II. When a, b, c are rational numbers, a 0: If = b² -4 a c = 0, then roots are rational and equal. If = b² -4 a c > 0, and is a perfect square of a rational number, then roots are rational and unequal. If = b² -4 a c > 0 but is not a square of rational number, then roots are irrational and unequal. They form a pair of irrational conjugates p +q, p - q where p, q Q, q> 0. If = b² -4 a c < 0, then roots are a pair of complex conjugates. Illustrative Examples Example: Discuss the nature of the roots of the following equations: (i) 4 x² -12 x +9 = 0 (ii) 3 x² -10 x +3 = 0 (iii) 9 x² -2 = 0 (iv) x² +x +1 = 0 Solution Here coefficients are rational, and discriminant = b² -4 a c = (-12)² -4 (4)(9) = 144 -144 = 0. Hence the roots are rational and equal. Here coefficients are rational, and discriminant = b² -4 a c = (-10)² -4 (3)(3) = 100 -36 = 64. Now = 64 > 0, and 64 is a perfect square of a rational number. Hence the roots are rational and unequal. Here coefficients are rational, and discriminant = b² -4 a c = (0)² -4 (9)(-2) = 72. Now = 72 > 0 but is not a perfect square of a rational number. Hence the roots are irrational and unequal. Here coefficients are rational, and discriminant = b² -4 a c = (1)² -4 (1)(1) = -3 < 0. Hence the roots are a pair of complex conjugates. Example Discuss the nature of the roots of the equation (m +6) x² +(m +6) x +2 = 0 Solution Discriminant = (m +6)² -4 (m +6)(2) = m² +12 m +36 -8 m -48 = m² +4 m -12 = (m +6)(m -2) Roots are real and equal if = (m +6)(m -2) = 0 i.e. if m = 2 (Ignoring m= -6, as then equation becomes 2=0) Roots are real and unequal when = (m +6)(m -2) > 0 i.e. when m < -6 or when m > 2 Roots are a pair of complex conjugates when = (m +6)(m -2) < 0 i.e. when -6 < m < 2 Exercise Find the nature of roots of the following equations without solving them: (i) x² +9 = 0 (ii) 4 x² -24 x +35 = 0 (iii) x² -22 x +1 = 0 (iv) 2 x² -25 x +3 = 0 Show that roots of the equation (x -a)(x -b) = a b x² where a, b R are always real. When are they equal? [Hint. = (a -b)² +(2 a b)²]. Show that the roots of the equation (x -a)(x -b) +(x -b)(x -c) +(x -c)(x -a) = 0, where a, b, c R are always real. Find the condition that the roots may be equal. What are the roots when this condition is satisfied? [Hint. = 2 ((a -b)² +(b -c)² +(c -a)²)] Discuss the nature of roots of the following equations: (i) 3 x² -2 x -3 = 0 (ii) x² -(p +1) x +p = 0 (iii) (x -a)(x -b) = a b. It is given that p Q, and a, b R. Find m so that roots of the equation (4 +m) x² +(m +1) x +1 = 0 may be equal. Show that the roots of the equation x² +2 (3 a +5) x +2 (9 a² +25) = 0 are complex unless a = 5/3. If a, b, c, d R show that the roots of the equation (a² +c²) x² +2 (a b +c d) x + (b² +d²) = 0 cannot be real unless they are equal. Determine a positive real value of k such that both the equations x² +k x +64 = 0 and x² -8 x +k = 0 may have real roots. Answers 1. (i) pair of complex conjugates (ii) rational and unequal (iii) real and unequal (iv) pair of complex conjugates 2. Roots are real and equal when a = b = 0 3. Roots are equal when a = b = c. Then roots are a, a 4. (i) real and distinct (ii) rational and distinct when p 1; when p = 1, roots are rational and equal (iii) real and distinct when a +b 0; when a +b = 0, roots are both 0. 5. 5, -3 8. k = 16 Sum and Product of Roots sum of roots: -b/a product of roots: c/a As you can see from the work below, when you are trying to solve a quadratic equations in the form of . The sum and product of the roots can be rewritten using the two formulas above. Sum and Product Formula Derived Example 1 The example below illustrates how this formula applies to the quadratic equation . As you, can see the sum of the roots is indeed and the product of the roots is . Picture of sum and product of roots formula Example 2 The example below illustrates how this formula applies to the quadratic equation x2 - 2x - 8. Again, both formulas--for the sum and the product boil down to -b/a and c/a, respectively. Picture of sum and product of roots formula
Posted on: Tue, 29 Jul 2014 06:02:46 +0000
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