parabolas

- x^2 -x + 3    = 2x^2 -1      re-arrange

0 = 3x^2 +x-4             use quadratic formula to find the x values where the graphs intersect

                                           a = 3     b   = 1      c = -4

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

                                          use the x values in the equations to calculate the corresponding 'y' values

(actually....you will only need the 'x' values  which will be   c    and a  )

                                                ...then you can answer the rest of the question

The parabolas defined by the equations y = -x^2 - x + 3 and y = 2x^2 - 1 intersect at points (a, b) and (c, d), where c is greater than or equal to a. What is c - a? Express your answer as a common fraction.

Hello Guest!

$y = -x ^ 2 - x + 3\\ y = 2x ^ 2 - 1\\ \color{black}2x^2-1=-x^2-x+3$

$3x^2+x-4=0$

$x = {-1 \pm \sqrt{1+4\cdot 3\cdot 4} \over 2\cdot 3}\\ x=\frac{-1\pm 7}{6}$

$a=-\frac{4}{3}\\ \color{black}b=\frac{23}{9}=2.5\overline {5}\\ c=1\\ d=1$

$c-a=1-(-\frac{4}{3})$

$c-a=\frac{7}{3}$

  !