Question

Please help I will give out brainliest

Answer 1

Answer:

i)x=0,x=1

ii)

$\frac{2}3} + \frac{\sqrt{40} }{6} \\\frac{2}3}-\frac{\sqrt{40} }{6}$

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:

$\frac{-(-4)+ or -\sqrt{(9-4)^2)-4(-2)(3)} }{2(3)}$

from there the equation gets simplified down to:

$\frac{4+ or -\sqrt{16-4(-6)} }{6}$

simplifying the formula even further gives us:

$\frac{4+ or -\sqrt{40} }{6}$

if we simplify the equation one last time, we get:

$\frac{2}{3} + or -\frac{\sqrt{40} }{6}$

therefore, the solutions to this equation are:

$\frac{2}{3} +\frac{\sqrt{40} }{6}\\ \frac{2}{3} - \frac{\sqrt{40} }{6}$