Answer:
c.7 and d.5
Step-by-step explanation:
Given the following equation :
[tex]x^{2}-6x+40=6x+5[/tex]
This equation is equivalent to the following equation :
[tex]x^{2}-6x+40=6x+5[/tex]
[tex]x^{2}-6x+40-6x-5=0[/tex]
[tex]x^{2}-12x+35=0[/tex]
Given an equation [tex]ax^{2}+bx+c=0[/tex] we can find the solutions using the quadratic equation (x1 and x2 are the solutions) :
[tex]x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]
Using this equations :
[tex]x^{2}-12x+35=0[/tex]
[tex]a=1\\b=-12\\c=35[/tex] ⇒
[tex]x1=\frac{-(-12)+\sqrt{(-12)^{2}-4.(1).(35)}}{2.(1)}=7[/tex]
[tex]x2=\frac{-(-12)-\sqrt{(-12)^{2}-4.(1).(35)}}{2.(1)}=5[/tex]
Given the two solutions x1 and x2 we can write the equation :
[tex]ax^{2}+bx+c=0[/tex] ⇒ [tex]a.(x-x1).(x-x2)=0[/tex] ⇒
[tex]x^{2}-12x+35=0[/tex] is equivalent to
[tex](x-7).(x-5)=0[/tex]
The solutions for this equation are 7 and 5.
Therefore, c.7 and d.5 are the solutions for this exercise.
