How can I solve this system of equation?1.9. The program should take three numbers: a; b; c and find the roots of the quadratic equation in the form: a x 2 + b x + c = 0If the value of the determinant of the quadratic equation is negative (i.e. △ &lt; 0), the program should write an appropriate message.1.10. Modify the 1.9 so that the parameters A and a are functions of the parameter ϕ, e.g. A ( ϕ = A ( 1.0 + 0.25 sin ⁡ ( ϕ ) ) a ( ϕ ) = a ( 1.0 + 0.5 ⋅ | sin ⁡ ( ϕ ) | )In the program, declare functions for the above equations and save the results in an array.I have solved the 1.9 as that was easy. But I am stuck with the following: (1) f ( x ) = f ( 1 + 0.25 sin ⁡ ( x ) ) (2) g ( x ) = g ( 1 + 0.5 | sin ⁡ ( x ) | ) My attempt:Given, f ( x ) = f ( 1 + 0.25 sin ⁡ ( x ) )Now, f ( 1 + 0.25 sin ⁡ ( x ) ) = a 16 sin 2 ⁡ x + ( a 2 + b 4 ) sin ⁡ x + ( a + b + c )Let, sin ⁡ x = ySo, the above equation becomes: y = − ( 2 a + b ) ± b 2 − 4 a c 2 a Hence, sin ⁡ x = − ( 2 a + b ) ± b 2 − 4 a c 2 a x = sin − 1 ⁡ ( − ( 2 a + b ) ± b 2 − 4 a c 2 a ) Are this procedure and solution correct?

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How can I solve this system of equation?1.9. The program should take three numbers: a; b; c and find the roots of the quadratic equation in the form:  a      x    2    +  b  x  +  c  =  0If the value of the determinant of the quadratic equation is negative (i.e.   △  &amp;lt;  0), the program should write an appropriate message.1.10. Modify the 1.9 so that the parameters A and a are functions of the parameter   ϕ, e.g.  A  (  ϕ  =  A  (  1.0  +  0.25  sin  ⁡  (  ϕ  )  )  a  (  ϕ  )  =  a  (  1.0  +  0.5  ⋅      |    sin  ⁡  (  ϕ  )      |    )In the program, declare functions for the above equations and save the results in an array.I have solved the 1.9 as that was easy. But I am stuck with the following:                    (1)                    f        (        x        )        =        f        (        1        +        0.25        sin        ⁡        (        x        )        )                                (2)                    g        (        x        )        =        g        (        1        +        0.5                  |                sin        ⁡        (        x        )                  |                )            My attempt:Given,  f  (  x  )  =  f  (  1  +  0.25  sin  ⁡  (  x  )  )Now,  f  (  1  +  0.25  sin  ⁡  (  x  )  )  =      a    16        sin    2    ⁡      x    +      (          a      2        +          b      4        )    sin  ⁡      x    +  (  a  +  b  +  c  )Let,   sin  ⁡      x    =  ySo, the above equation becomes:   y  =            −      (      2      a      +      b      )      ±                        b          2                −        4        a        c                    2      a      Hence,                    sin        ⁡                  x                                    =                                            −              (              2              a              +              b              )              ±                                                b                  2                                −                4                a                c                                                    2              a                                                          x                            =                  sin                      −            1                          ⁡                              (                                                            −                  (                  2                  a                  +                  b                  )                  ±                                                            b                      2                                        −                    4                    a                    c                                                                    2                  a                                                      )                              Are this procedure and solution correct?

hamputlnf · Accepted Answer

Step 1I think that Problem 1.10 is almost as simple as Problem 1.9. The only difference is that the parameters a and b or c (there is no parameter A in Problem 1.9, so this should be a misprint) instead of constant values are functions of the parameter   φ declared in the program (so a program solving Problem 1.10 can be a small modification of the program solving Problem 1.9). A solver has the freedom to choose the functions.Step 2The equality   A  (  ϕ  )  =  A  (  1.0  +  0.25  sin  ⁡  (  ϕ  )  ) is not an equation to solve, it is an example of a function of a parameter   φ. A bit confusing here is that A from the left-hand side is a name of a function and the left-hand side equals to value of A of the argument   φ, whereas A from the right side is a constant and the right-hand side is a product of this A by   1.0  +  0.25  sin  ⁡  (  ϕ  ). Similarly for the other equality.

How can I solve this system of equation? 1.9. The program should take three numbers: a; b; c and find the roots of the quadratic equation in the form: ax^2+bx+c=0.

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