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

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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 Answers 3

up vote 2 down vote accepted

Some easy ideas to attack your problem with integer solutions:


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):


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

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

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!

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Thanks...for your solution..but seriously not getting this way if you can simplify more make it easier for me in understanding.. – Umair Shah Mar 24 '14 at 13:21
Which part is confusing you? I just made some edits . – Lee Yiyuan 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. – Lee Yiyuan Mar 24 '14 at 13:30
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

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?

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Thanks...for your solution..but seriously not getting this way if you can simplify more 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 please if you solve it to the will be very kind of you..please – Umair Shah Mar 24 '14 at 13:27
@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 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 I hope that you may not mind....please – Umair Shah Mar 24 '14 at 14:14

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