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I have one question please...I solved it in this way..so I am not sure..that is it right or not? and If I solved it in the wrong way..so would like to know about the correct way and method to solve it..so please :

Solve the equation (x-3) (x+9) (x+5) (x-7) = 385

and here is I solved it in this way : 

Solve the equation:
(x-3) (x+9) (x+5) (x-7) = 385
x ( (1-3) (1+9) (1+5) (1-7)) = 385
x ( (-2) (10) (6) (-6) ) = 385
x ( (-20) (6) (-6) ) = 385
x ( (-120) (-6) ) = 385
x ( 720 ) = 385
720x = 385
x = 385/720
x = 0.5347222

waiting for your replies...

mle
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Umair Shah
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  • Good work on attempting a solution. You can learn a lot from experimenting.

    With regards to your attempt, when you take x out as a common factor, you have to take x out of all components.

    So (x-3) = x(1 - 3/x) A good way to test your steps is to expand it out again!

     – Fuzz Mar 24 '14 at 14:04
  • @Fuzz Thanks..for encouragment..I quitely do it often..whenever I don't understand..still I solve some way..but not sure..that what I did is in the right way or in the wrong way? – Umair Shah Mar 24 '14 at 14:11

3 Answers3

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Let $y = x + 1$ (I got this by taking the average of the four). Then the problem becomes:

$$(y-4)(y+8)(y+4)(y-8) = 385$$

which can be simplified nicely (by multiplying out similar terms) into:

$$\begin{align}(y^2 - 16)(y^2 - 64) &= 385 \\&= 7\cdot55\\&=-7\cdot-55\end{align}$$

The difference between $7$ and $55$ is $48$. But this is exactly the difference between $y^2 - 16$ and $y^2 - 64$. Hence we equate them according to their magnitudes ($y^2 - 16 = 55$) to deduce that:

$$\begin{align}&y^2 - 16 = 55 \\\implies&y^2 = 71\\\implies &(x+1)^2 = 71 \\ \implies &x = -1\pm\sqrt{71}\end{align}$$

Now, we deal with $-7$ and $-55$ in a similar manner:

$$\begin{align}&y^2 - 16 = -7 \\\implies&y^2 = 9\\\implies &(x+1)^2 = 9 \\ \implies &x = -1\pm3\end{align}$$

We conclude with with our four solutions for $x$:

$$x = -1\pm\sqrt{71}, -1\pm3$$

The best part: No quadratic formula used!

Yiyuan Lee
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  • Thanks...for your solution..but seriously not getting this way if you can simplify more please..to make it easier for me in understanding.. – Umair Shah Mar 24 '14 at 13:21
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    Which part is confusing you? I just made some edits . – Yiyuan Lee Mar 24 '14 at 13:22
  • Great...Thanks...for solving it to me... – Umair Shah Mar 24 '14 at 13:29
  • Glad I could help. Since, you are new (i presume) to Math.SE: Do remember to mark any answer posted to your questions if you find it useful and sufficient, via the tick beside it. – Yiyuan Lee Mar 24 '14 at 13:30
  • Thanks..@Yiyuan Lee ..for your quite good way to solve it..but rather as it is in short questions so I will take the easy way to solve it...and this way seems little longer..so I hope that you may not mind....please – Umair Shah Mar 24 '14 at 13:50
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    Neat! You might like my explanation of how I came to discover something like this at my sci.math posts Some turbo-charged high school algebra 21 August 2008 and 25 August 2008 (which I've since learned is a somewhat standard technique). – Dave L. Renfro Mar 24 '14 at 13:52
  • @DaveL.Renfro Thanks for the references, they seem nice! – Yiyuan Lee Mar 24 '14 at 13:57
  • @UmairShah No problem, I would agree that DonAntonio's solution is indeed a lot faster and simpler in this case. I would have marked his as the answer as well if I were you. – Yiyuan Lee Mar 24 '14 at 13:58
  • @YiyuanLee Thank you so much for your favour...! I will look after for more problems...I like your way...of solving problems..once again thanks... – Umair Shah Mar 24 '14 at 14:05
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Observe that $\displaystyle -3+5=-7+9$

So, $$(x-3)(x+5)=x^2+2x-15, (x+9)(x-7)=x^2+2x-63$$

Set $\displaystyle x^2+2x=u$ to find $\displaystyle (u-15)(u-63)=385$

$\displaystyle\implies u^2-78u+560=0$

Using Quadratic Equations Formula, $\displaystyle u=8,60$

or using Middle Term Factor, $\displaystyle u^2-78u+560=u^2-(70+8)u+70\cdot8=(u-70)(u-8)$

So, we have $\displaystyle x^2+2x=8$ or $\displaystyle x^2+2x=70$

Can you take it from here?

lab bhattacharjee
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  • Thanks...for your solution..but seriously not getting this way if you can simplify more please..to make it easier for me in understanding. – Umair Shah Mar 24 '14 at 13:22
  • @UmairShah, please find the edited version? – lab bhattacharjee Mar 24 '14 at 13:25
  • Sure...thanks...really...getting this way...now please if you solve it to the end...so will be very kind of you..please – Umair Shah Mar 24 '14 at 13:27
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    @UmairShah, If you have understood the solution for $u,$ you should be able to find $x$ in each case – lab bhattacharjee Mar 24 '14 at 13:29
  • Thanks for your answer....you have solved in a quite good way..but rather as it is in short questions so I will take the easy way to solve it...and this way seems little longer..so I hope that you may not mind....please – Umair Shah Mar 24 '14 at 14:14
  • @UmairShah, acceptance or voting should not be the concern here. What should matter most is sharing knowledge, rectifying one's mistake/ misconception(s). – lab bhattacharjee Mar 24 '14 at 16:27
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Some easy ideas to attack your problem with integer solutions:

$$(x-3)(x+9)(x+5)(x-7)=385=5\cdot7\cdot11\implies$$

one of the factors in the LHS must be $\;\pm1\;$ , and after a little action we can see this happens for $\;x=2\;$ since the signs fit nicely (two positive, two negative):

$$(2-3)(2+9)(2+5)(2-7)=(-1)\cdot11\cdot7\cdot(-5)=385$$

Try now also $\;x=-4\;$ ...

DonAntonio
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  • Great...very nice..and very easy way..to solve it..Thanks..I like it...And very easy method to solve the question... – Umair Shah Mar 24 '14 at 13:25