+0  
 
0
40
1
avatar

3x^2-4x+8=8x^2-5x-3

Guest Aug 23, 2017
Sort: 

1+0 Answers

 #1
avatar+6901 
0

3x^2-4x+8=8x^2-5x-3

 

\(3x^2-4x+8=8x^2-5x-3\\ 4x^2-x-11=0\)    8 - 3 = 5  not  4 !!

a        b       c

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

\(x = {1 \pm \sqrt{1-4\cdot 4\cdot (-11)} \over 2\cdot 4}\)

\(x=\frac{1\pm \sqrt{177}}{8}=0.125\pm 1.6630168\)

 

\(x_1=1.7880168\\ x_2=-1.5380168\)

 

This result is correct for (\(4x^2-x-11=0\)),

but not for (\(3x^2-4x+8=8x^2-5x-3\)).

Why ?

8 - 3 = 5     not  4 !!

Thanks for the help Alan and heureka!
I will correct immediately.

\(3x^2-4x+8=8x^2-5x-3\\ 5x^2-x-11=0\)

a        b       c

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

\(x = {1 \pm \sqrt{1-4\cdot 5\cdot (-11)} \over 2\cdot 5}\)

\(x=\frac{1\pm\sqrt{221}}{10}=0.1\pm 1.48660687 \)

 

\(x_1=1,58660687\)

\(x_2=-1.38660687\)

 

laugh  !

asinus  Aug 23, 2017
edited by asinus  Aug 23, 2017
edited by asinus  Aug 23, 2017
edited by asinus  Aug 23, 2017

9 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details