Michael M. answered 10/19/20
Struggling with STEM classes? I can help!
Okay, Sophia, let's see if we can0 figure this out...
First equation: y=3x^2
Second equation: y=3x-2
If the first equation AND the second equation both equal y (on the left side), then the first equation AND the second equation are equal to one another. So now we have 3x^2=3x-2. Graphically, this indicates that we are looking for the intersection (if there is one) of a parabola and a line.
Now, let's subtract 3x-2 from both sides of this equation, which will give us 3x^2-(3x-2)=3x^2-3x+2=0
This is a quadratic equation, which can be solved using the quadratic formula: For ax^2+bx+c=0, we have
x={-b+-sqrt[b^2-4ac]}/2a (here +- means "plus or minus" and sqrt means "square root")
For our quadratic equation, a=3, b=-3 and c=2. Substituting values for a, b and c, we have
x={-(-3)+-sqrt[(-3)^2-4(3)(2)]}/2(3), or x={3+-sqrt[9-24]}/6
Given that a,b and c are real numbers, there are three possibilities:
b^2-4ac is positive, in which case the roots are real and unequal
b^2-4ac is zero, in which case the roots are real and equal
b^2-4ac is negative, in which case the roots are imaginary and unequal.
Notice that b^2-4ac is 9-24, which is -15 and negative; therefore, there are TWO SOLUTIONS to this equation, both of which are imaginary, and the two solutions are unequal; or, put a different way, there are no real solutions to this problem (because the solutions are imaginary).
I hope this helps.