+0  
 
0
138
1
avatar

The x-intercetps of the parabola y = 2x^2 + 13x - 4\(+ 2x^2 - 6x + 5\) are at (p,0) and (q,0). Find p and q. Enter your answer in the form "p,q".

 Jan 16, 2022
 #1
avatar+13900 
0

Find p and q.

 

Hello Guest!

 

\( y = 2x^2 + 13x - 4+ 2x^2 - 6x + 5\\ y=4x^2+7x+1\\ y=x^2+1.75x+0.25=0 \)

\(x=-0.875\pm \sqrt{0.515625}\)

\(\{p,q\}=\{-1.59307033082,-0.156929669183\}\)

laugh  !

 Jan 17, 2022

22 Online Users

avatar
avatar