Answer:
First image attached
The error was done in Step E, because student did not multiply [tex]2\cdot x -8[/tex] by the negative sign in numerator. Step E must be [tex]\frac{2\cdot x -5}{(x+4)\cdot (x-4)}[/tex].
Second image attached
The error was done in Step C, because the student omitted the [tex]2\cdot a \cdot b[/tex] of the algebraic identity [tex](a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}[/tex]. Step C must be [tex]5\cdot x = x^{2}+4\cdot x + 4[/tex]
Step-by-step explanation:
First image attached
The error was done in Step E, because student did not multiply [tex]2\cdot x -8[/tex] by the negative sign in numerator. The real numerator in Step E should be:
[tex]3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x[/tex]
Hence, Step E must be [tex]\frac{2\cdot x -5}{(x+4)\cdot (x-4)}[/tex].
Second image attached
The error was done in Step C, because the student omitted the [tex]2\cdot a \cdot b[/tex] of the algebraic identity [tex](a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}[/tex]. Step C must be [tex]5\cdot x = x^{2}+4\cdot x + 4[/tex]
And further steps are described below:
Step D
[tex]x^{2}-x+4 = 0[/tex]
Which according to the Quadratic Formula, represents a polynomial with complex roots. That is: ([tex]a = 1[/tex], [tex]b = -1[/tex], [tex]c = 4[/tex])
[tex]D = b^{2}-4\cdot a\cdot c[/tex]
[tex]D = (-1)-4\cdot (1)\cdot (4)[/tex]
[tex]D = -17[/tex] (Conjugated complex roots)
Step E
[tex](x-0.5-i\,1.936)\cdot (x-0.5+i\,1.936) = 0[/tex]
Step F
[tex]x = 0.5+i\,1.936\,\lor\,x = 0.5-i\,1.936[/tex]
