ALGEBRA - solved problems
  Polynomial and/or polynomial functions and equations
Quadratic equations with absolute value
141.
    Solve the absolute value equation  | x2 - 3 | = - x + 3.
Solution:   Using the definition of the absolute value of a function we can write,
if    x2 - 3 > 0   then   | x2 - 3 | = x2 - 3,
 and given equation   x2 - 3 = - x + 3  or
          x2 + x - 6 = 0.   By factoring the equation,
x2 + 3x - 2 x - 6 = [x(x + 3) - 2(x + 3)] 
          x2 + x - 6 = (x + 3)(x - 2) = 0
so,  x + 3 = 0x1 = - 3  and  x - 2 = 0,  x2 = 2.
If    x2 - 3 < 0   then   | x2 - 3 | = - (x2 - 3),
and given equation   - ( x2 - 3) = - x + 3  or
- x2 + x = x(- x + 1) = 0
 so,  x3 = 0  and  - x + 1 = 0  x4 = 1.
Therefore, the solutions of the given equation,  x1 = - 3,  x2 = 2x3 = 0  and  x4 = 1 are the abscissas of the intersection points of,  = | x2 - 3 | and  = - x + 3  functions, as shows the above figure.
142.
    Solve the absolute value equation  | 2x2  - x - 3 | = - x + 5.
Solution:   Using the definition of the absolute value of a function we can write,
if         2x2  - x - 3 > 0 
then    | 2 x2  - x - 3 | = 2 x2  - x - 3,
and given equation  2 x2  - x - 3 = - x + 5
                  or       2x2 = 8,     x2 = 4
 so,   x1,2 = + 2,    x1 = - 2  and  x2 = 2.
if         2 x2  - x - 3 < 0 
then    | 2x2  - x - 3 | = - (2x2  - x - 3),
and given equation   - (2 x2  - x - 3) = - x + 5,
      - 2x2  + 2x - 2 = 0   or   x2  - x + 1 = 0,
 thus, there are no real solutions in this case.
Therefore, the solutions of the given equation,  x1 = - 2  and  x2 = 2 are the abscissas of the intersection points of,  = | 2 x2  - x - 3 | and  = - x + 5  functions, as is shown in the above figure.
Cubic functions or the third degree polynomial
y = a3x3 + a2x2 + a1x + a0     or      y - y0 = a3(x - x0)3 + a1(x - x0),
By setting  x0 = 0  and  y0 = 0 we get the source cubic function  y = a3x3 + a1x  where  a1= tan at .
Coordinates of the point of inflection coincide with the coordinates of translations, i.e.,  I(x0, y0).
143.
    Find the coordinates of translations, the zero point, the point of inflection and draw graphs of the
cubic function  y = - x3 + 3x2 -  5x + 6 and its source function.
Solution:   1)  Calculate the coordinates of translations
         y0 f(x0)   =>    y0 = f (1) = -13 + 3 12 - 5 1 + 6  = 3,                   y0 = 3.
Therefore, the point of inflection  I(1, 3).
2)  To get the source cubic function, plug the coordinates of translations into the general form of the cubic,
y + y0 =  a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
thus,      y + 3 = -1 (x + 1)3 + 3 (x + 1)2 - 5 (x + 1) + 6    =>    y = - x3 - 2the source function.
Since given function is symmetric to its point of inflection, and as the y-intercept a0 = 6, then the x-intercept or zero of the function must be at the point (2, 0).
Therefore,   - x3 + 3x2 -  5x + 6 = - (x - 2)(x2 - x + 3).
Equations with rational expressions
Rational equations - Linear equations
144.
    Solve the following rational equations.
Solutions: 
 
 
   
   
Rational equations - Quadratic equations
145.
    Solve the following rational equations.
Solutions: 
 
 
Solved problems contents
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