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Analytical Toolbox

Analytical Toolbox. Higher order Equations and Solutions By Dr J.P.M. Whitty. Learning objectives. After the session you will be able to: Find the solution to Quadratic equations using graphical means. Find algebraic solutions to quadratic equations.

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Analytical Toolbox

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  1. Analytical Toolbox Higher order Equations and Solutions By Dr J.P.M. Whitty

  2. Learning objectives After the session youwill be able to: Find the solution to Quadratic equations using graphical means. Find algebraic solutions to quadratic equations. Solve simultaneous equations one quadratic and one linear. Use graphical techniques to estimate the solution of higher order equations Use math software to solve systems of equations.

  3. Learning Check: Solve simultaneous equations Solve the following simultaneous equations y=2x+4 & 3x=7-5y using three different mathematical techniques, i.e: Algebraically Graphically MATLAB matrix method

  4. Solution -2x+y=4 3x+5y=7 -6x+3y=12 6x+10y=14 13y=26; y=2 -2x+2=4 x=(4-2)/-2=-1 Commands >> ezplot('-2*x+y=4') >> hold on >> ezplot('3*x+5*y=7') >> grid on >> ginput

  5. Solution cont… The MATLAB matrix method is as follows: A=[-2 1; 3 5] b=[4; 7] A\b ans=[-1 2]

  6. Recap: Linear Functions We have already learned about linear functions, for example: f(x) =y= 4x + 1 g(x) =y= -6x -2 h(x) =y= (-2/3)x + 2 The graphs of these functions are straight lines and each equation gives us “clues” as to what each line looks like: f(x) = 4x + 1 Slope of 4

  7. Quadratic Functions What do the functions f(x) = 2x2 + 4x – 3 g(x) = x2 – 4x h(x) = -x2 – 5x + 6 have in common? They are all QUADRATIC FUNCTIONS that can be written in standard form: f(x) = ax2 +bx + c where a ≠ 0 Pay close attention… these coefficients a, b and c will be cluesthat help us to know what the graphs of these quadratic functions look like…

  8. Graphs of Quadratic Functions Quadratic Functions graph into a shape called a Parabola If the coefficient of the x2 term, “a”, is negative, the parabola opens downward (Sad) If the coefficient of the x2 term, “a”, is positive, the parabola opens upward (Happy) f(x) = -x2 + 5 Maximum Point f(x) = x2 - 4 Vertex Minimum Point

  9. The parabola The graph of is a parabola with vertex (0, 0) Example: Locate the vertex and find the axis of symmetry of the following parabola. Does it open up or down? Graph the parabola.

  10. Quadratic equations Any equation that can be written so that one side is a quadratic and the other side is zero is called a quadratic equation. In fact, we can consider a “general form” of quadratic equation. However a quadratic is given to us, we can always rewrite it as:

  11. Solving Quadratics As with all equations, solving them simply means finding the values that make them true. The easiest quadratic equation we could be asked to solve would be something like: There are TWO values that make this true: x = 2 and x = -2 (in MATLAB roots([1 0 -4]); the value of the coefficients). We call these the roots of the equation. In fact this idea of two solutions extends to just about all quadratic equations.

  12. Solution by factorization At GCSE level you would have spent a lot of time practising factorization of quadratic expressions. Here’s why... Consider the following equation: We can factorise the LHS of the equation as follows: This leaves us with the equation looking like this:

  13. Sketching the curve Here the solution to renders intercepts on the x-axis, while putting x=0; gives the intercept on the y-axis, hence generating the required three points, thus: For a forth point let x=1.5; y=0.25

  14. Try one: Solve the equation hence or otherwise sketch the curve: Factorize: x=0;y=-3 x=1;y=-4

  15. Formula solution to quadratic equations The process of completing the square leads us to a formula for solving any quadratic equation. We can solve a quadratic in the form: With the formula: Those feeling powerful should try to derive this formula. The starting point should be a general formula for completing the square.

  16. The formula for quadratic equations Examples

  17. Example solutions: Examples:

  18. The formula for quadratic equations Note that you MUST rearrange an equation into ax2 + bx + c = 0 form before using the formula. Also… remember that NOT every quadratic equation has two solutions, and indeed, some have none at all!! We’ll look at the following examples with the formula.

  19. Repeated roots What’s going on here is that the expression x2 – 6x + 9 is a perfect square. Namely (x – 3)2. So we only have one solution to x2 – 6x + 9 = 0. We say that 3 is a repeated root of the equation x2 – 6x + 9 = 0 .

  20. No Solutions?? Our problem here is that we CANNOT find the square root of a negative number. In situations like this, we simply say that the equation has no solutions. That is, there are NO values of x that make 3x2 + 2x + 20 equal 0.

  21. A general approach for solving quadratics Rearrange into the form Try to factorise. The idea is that you are so well practised in factorisation that this shouldn’t take too long! If you can’t factorise, use the formula. You’ll note that completing the square isn’t mentioned. That’s why we didn’t spend long talking about it!! The reasons I bothered to include it at all are (a) the method leads to the formula, and (b) so that if anyone else ever mentions it to you, you will know what they mean, and have somewhere to come back to!

  22. When is a quadratic not a quadratic? We will often be faced with equations that don’t look very quadratic, that turn out to be when rearranged. All of these can be rearranged-to and/or tackled as quadratics.

  23. When it’s a jar…

  24. Summary You should now know what is meant by an equation know what is meant by a quadratic equation be able to solve a quadratic equation by factorisation be able to rearrange and solve without losing a solution be able to use the completing the square technique know and be able to use the formula for quadratic equations be able to recognise (thanks to the formula) when a quadratic equation has no solutions

  25. Examination Type Question Using suitable math software or otherwise sketch the graph of the following functions f(x)=3x2+2x-7 [6] g(x)=2x+3 [4] From the graph obtain approximate values of the interception points [2]. Use an analytical method to obtain better solutions [5]. How could you obtain better approximations using math software [3] [20 marks]

  26. Solution: Use the MATLAB ginput command to obtain approximate solutions ans = -1.7807 -0.6651 1.8386 6.9621

  27. Analytical solutions: f(x)=3x2+2x-7 g(x)=2x+3 *Better solutions could be obtained from MATLAB ginput command or alternatively as for the roots of the quadratic

  28. Further Examination Type Q: The graph shows two functions namely: 4x2-2x+1 & x+4. Explain which curve is which giving reasons for your answer [5]; and find the interception points [5]. [10 marks]

  29. Solution The first function is a parabolaB2 (only curveB1) due to the polynomial being of second orderB1. The second graph is a is a lineB1 as it has the form y=mx+cB1. M1 M2 A1 A1

  30. Another Exam Question Find the interception of the graphs: You could use the formula here also, with a=3, b=-6 and c=0 M1A1 and [5] M1 M1A1 A1 A1

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