Question

solve this as quadratic form : (x+1)(2x+3)(2x+5)(x+3)=945

Answer 1

I would first try to write 945 as a product of its prime factors.

Answer 2

can you do that?

Answer 3

it is to be solved like (x+a)(x+b)(x+c)(x+d) where the sum of any two from a,b,c,d would be equal to other two...

Answer 4

you need to find 4 numbers that, when multiplied together, will give you 945 - correct?

Answer 5

those 4 numbers are:
1. x+1
2. 2x+3
3. 2x+5
4. x+3

Answer 6

in problems like this, it is usually best to start by breaking down 945 into its prime factors

Answer 7

for example :
(x+2)(x+3)(x+4)(x+1)=2
[(x+2)(x+3)][(x+4)(x+1)] =2 (in this group i have 2 product of square braces in which i have grouped 2+3 , 4+1 = 5
the same have to be done somehow in this question!

Answer 8

sorry but I do not understand what you are trying to achieve. I can solve this and find a value for x such that (x+1)(2x+3)(2x+5)(x+3)=945 by breaking 945 down into its prime factors.

if you are trying to do something different, then I do not quite understand what it is that you are trying to do.

maybe someone else will be able to help you better.

Answer 9

okay can you plz do it your way?

Answer 10

I can certainly guide you to the solution.

so first step is to break 945 into its prime factors.

what do you get for that?

Answer 11

3*3*3*5*7

Answer 12

good - now we need to group this into a product a 4 numbers - can you see how to group them?

Answer 13

*product of

Answer 14

uhm like 3*3*3*5*7=9*3*5*7

Answer 15

yup - so the smallest of these is 3 - agreed?

Answer 16

yup!

Answer 17

and the smallest in the product: (x+1)(2x+3)(2x+5)(x+3)
is (x+1)

so I would set: x + 1 = 3

and solve for x

Answer 18

x=2 ?

Answer 19

yup! - now try substituting that into the equation and see if you get the right values

Answer 20

i have three answers given in my text book : 2, -6 and one more value.. like complex no.

Answer 21

hmmm - I have never really come across a problem like this before and the way you described the solution above is not something I am familiar with.

is there a specific name for this type of problem?

Answer 22

I'd be interested in learning about it.

Answer 23

yup .. first yo have to convert them into quadratic form.. then solve.

Answer 24

so are you saying that you have to form the product of two quadratics, and then find all possible values for each one of those quads and then solve them?

Answer 25

is there a simpler example that you have solved that you can show me here - maybe then I will understand better?

Answer 26

here's a similar question : http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.126457.html

Answer 27

wait - that uses the same method I used O.O

Answer 28

but in my text book it has converted into quadratic form , then solved.

Answer 29

can i show you the solution of this one.. ?

Answer 30

yes please - that may help to explain the process

Answer 31

SOLUTION: (x+1)(x+2)(x+3)(x+4) = 120
[ (x+1)(x+4) ] [(x+2)(x+3)] =120
(x^2+5x+4) (x^2+5x+6)=120
put y=x^2+5x

Answer 32

oh I see now - let me see if I can try to solve this question in a similar fashion

Answer 33

you can the two two constants in the square braces add to give the same value thats the idea... like in first brace (4+1=5) , in second (3+2=5). So that's the arrangement.

Answer 34

*you can see

Answer 35

I believe you need to group them as follows:
(x+1)(2x+3)(2x+5)(x+3) = (x+1)(x+3) * (2x+3)(2x+5)

Answer 36

then you'll get:\[(x^2+4x+3)(4x^2+16x+15)\]

Answer 37

and you can set: \(y=x^2+4x\)

Answer 38

i have also given a hint for this ques. : LHS : 4[ (x+1) (x+3)] [ (x+3/2) (x+5/2) ]

Answer 39

so we end up with:\[(y+3)(4y+15)=945\]

Answer 40

and you can expand that and solve as a quadratic

Answer 41

\[4y^2+27y+45=945\]\[4y^2+27y-900=0\]\[(y-12)(4y+75)=0\]\[y=12\text{ or }-18\frac{3}{4}\]

Answer 42

so then you just use this to find x from the relation: \(y=x^2+4x\)

does that seem correct to you?

Answer 43

yes :D i got this thank you ... :D

Answer 44

thank YOU for teaching me a new technique! :)

Answer 45

:D

thank you too...