somehow it's hard to find the roots manually since
b^2 - 4ac <0
Solve 2x^2=3x+2 by graphing. If excat roots cannot be found, state the consecutive integers between which the roots are located.
I don't have an answer but I made a table and plugged in these values for x respectfully: -1/4, -2/4, -3/4, -1, 0
Thats all I have
2 answers
first of all bring all terms to one side
2x^2 - 3x - 2 = 0
now let f(x) = 2x^2 - 3x - 2
did you notice that it factors?
f(x) = (2x+1)(x-2)
so the x-intercepts are -1/2 and 2
which then become the roots of the equation from above.
2x^2 - 3x - 2 = 0
now let f(x) = 2x^2 - 3x - 2
did you notice that it factors?
f(x) = (2x+1)(x-2)
so the x-intercepts are -1/2 and 2
which then become the roots of the equation from above.
1 answer