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

 Aug 23, 2017
 #1
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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  !

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

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