Question #2b050

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Question #2b050

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Answer 1 · 2015-06-24T07:16:04

Square the both sides

Explanation:

$∣ ∣ x^{2} - 3 x ∣ ∣ = - 4 x + 6$ $|x^2-3x|=-4x+6$
$\sqrt{{(x^{2} - 3 x)}^{2}} = - 4 x + 6$ $sqrt((x^2-3x)^2)=-4x+6$
$x^{2} {(x - 3)}^{2} = {(- 4 x + 6)}^{2}$ $x^2(x-3)^2=(-4x+6)^2$
$x^{2} (x^{2} - 6 x + 9) = 16 x^{2} - 48 x + 36$ $x^2(x^2-6x+9)=16x^2-48x+36$
$x^{4} - 6 x^{3} + 9 x^{2} - 16 x^{2} + 48 x - 36 = 0$ $x^4-6x^3+9x^2-16x^2+48x-36=0$
$x^{4} - 6 x^{3} - 7 x^{2} + 48 x - 36 = 0$ $x^4-6x^3-7x^2+48x-36=0$
As sum of all coefficients are zero, x-1 is a factor.
$(x - 1) (x^{3} - 5 x^{2} - 12 x + 36) = 0$ $(x-1)(x^3-5x^2-12x+36)=0$
The rest of the cubic polynomial can be factored as $6 \cdot 2 \cdot - 3$ $6*2*-3$
So the roots are now $x = 1, 2, 6, - 3$ $x=1,2,6,-3$

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Question #2b050

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