+0

# How do I solve this equation?

0
974
5

$${ \sqrt{x-1}*(1-6x)-(1/ \sqrt{x-1}*2)*(x-3x^2) \over x-1}$$

Nov 5, 2015
edited by Namodesto  Nov 5, 2015

#4
+30

$$\dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\sqrt{x-1}}\cdot (x-3x^2) } { x-1 }$$

$$\small{ \begin{array}{rcll} &=& \left[ \dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { x-1 } \right] \cdot \dfrac{ 2\cdot \sqrt{x-1} } { 2\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ \sqrt{x-1}\cdot 2\cdot \sqrt{x-1}\cdot(1-6x)- \frac{2\cdot \sqrt{x-1}}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot \left(\sqrt{x-1}\right)^2 \cdot(1-6x)-(x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( x-6x^2-1+6x )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( 7x-6x^2-1 )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 14x-12x^2-2 - x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 13x-9x^2-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot (x-1)^{\frac12} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot(x-1)^{\frac32} } \\\\ \end{array} }$$ .
Nov 5, 2015
edited by heureka  Nov 5, 2015
edited by heureka  Nov 5, 2015

#1
+5

Hi Namodesto,

You cannot solve it because it is not an equation.

You can only simplify it

$${ \sqrt{x-1}*(1-6x)-(1/ \sqrt{x-1}*2)*(x-3x^2) \over x-1}\\ =\left[\sqrt{x-1}*(1-6x)-(\frac{2}{ \sqrt{x-1}})*(x-3x^2) \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\frac{x-6x^2-1+6x-2 x+6x^2}{\sqrt{x-1}*(x-1)} \\ =\frac{5x-1}{(x-1)^{3/2}} \\$$

.
Nov 5, 2015
#2
+5

Hi Melody,

http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1

I have problem understanding the last step. The following problem above was taken from an example of this website. How did he get to the final answer being -9x^2 + 13x - 2 from what I've written above?

Nov 5, 2015
#3
+5

Hi again Namodesto,

A part of the problem is that you presented the question incorrectly.  1/(sqrt(x-1)*2) the bottom needed to be in brackets.

You did not hve brackets which meant your 2 was up the top when it should  have been down the bottom.

I have been called away.  I will be back in about a half hour.  sorry.

I got t there answer except I got -2 anstead of +2

I will look more in a while :))

Must run .

Ok I am back and I can see that Heureka is already preparing another answer for you.

That +2 on  the first answer on this web page that you have since inserted  :-

http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1

is a typo, the answer further down is -2   and it is the same as I got when I redid your expression with the 2 down the bottom instead of up the top.  :)

I'll leave it now untill we both see what Heureka is preparing for you :))

Nov 5, 2015
edited by Melody  Nov 5, 2015
edited by Melody  Nov 5, 2015
#4
+30

$$\dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\sqrt{x-1}}\cdot (x-3x^2) } { x-1 }$$

$$\small{ \begin{array}{rcll} &=& \left[ \dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { x-1 } \right] \cdot \dfrac{ 2\cdot \sqrt{x-1} } { 2\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ \sqrt{x-1}\cdot 2\cdot \sqrt{x-1}\cdot(1-6x)- \frac{2\cdot \sqrt{x-1}}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot \left(\sqrt{x-1}\right)^2 \cdot(1-6x)-(x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( x-6x^2-1+6x )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( 7x-6x^2-1 )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 14x-12x^2-2 - x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 13x-9x^2-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot (x-1)^{\frac12} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot(x-1)^{\frac32} } \\\\ \end{array} }$$ heureka Nov 5, 2015
edited by heureka  Nov 5, 2015
edited by heureka  Nov 5, 2015
#5
+5

Thank you so much! Nov 5, 2015