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# help

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

Guest Aug 23, 2017
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### 1+0 Answers

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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

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

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

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

$$x_1=1,58660687$$

$$x_2=-1.38660687$$

!

asinus  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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