12.55 part b 6th edition

Moderators: Chem_Mod, Chem_Admin

pamcoronel1H
Posts: 45
Joined: Fri Sep 28, 2018 12:25 am

12.55 part b 6th edition

Postby pamcoronel1H » Sun Feb 10, 2019 10:17 pm

I am having a bit of trouble with the quadratic equation in this question. I got the quadratic equation of x^2+0.3x-0.06, but when I put this into the quadratic equation, I get a negative number in the square root part (sqrt(-.15)) and I can't take the square root of that. Anyone know what to do in this case?

Thank you!

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Re: 12.55 part b 6th edition

Postby Tam To 1B » Mon Feb 11, 2019 10:03 am

You should plug it in as -0.3+√(0.3^2 - 4(1)(-0.6)) and divide it all by 2. You should get a positive number inside the square root and x = 0.137.


Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”

Who is online

Users browsing this forum: No registered users and 2 guests