Answer:
i)x=0,x=1
ii)
[tex]\frac{2}3} + \frac{\sqrt{40} }{6} \\\frac{2}3}-\frac{\sqrt{40} }{6}[/tex]
Step-by-step explanation:
so I did not use the graphs actually.
for the answer to question i), i wrote out the equation as given:
3x^2-3x+2=2
then, i subtracted the two from the right side of the equation to make the equation set equal to 0, and I got:
3x^2-3x+2-2=+2-2, which becomes 3x^2-3x=0
then, I used the quadratic method of taking out the greatest common factor, which in this case is 3x, and the equation becomes:
3x(x-1)=0
finally, for this equation, I set 3x=0 and x-1 =0 to get the solutions to this equation, as shown below
3x=0(divide both sides by 3 to get)x=0
x-1=0(add 1 to both sides to get)x=1
now for the second equation, I had no other choice but to use the quadratic formula to find the solutions.
so I set up the quadratic formula as shown:
[tex]\frac{-(-4)+ or -\sqrt{(9-4)^2)-4(-2)(3)} }{2(3)}[/tex]
from there the equation gets simplified down to:
[tex]\frac{4+ or -\sqrt{16-4(-6)} }{6}[/tex]
simplifying the formula even further gives us:
[tex]\frac{4+ or -\sqrt{40} }{6}[/tex]
if we simplify the equation one last time, we get:
[tex]\frac{2}{3} + or -\frac{\sqrt{40} }{6}[/tex]
therefore, the solutions to this equation are:
[tex]\frac{2}{3} +\frac{\sqrt{40} }{6}\\ \frac{2}{3} - \frac{\sqrt{40} }{6}[/tex]
