finding numbers

**mark** · August 26th 2009, 07:52 AM

i have a question to do with quadratic equations but what i'm asking isn't really about quadratics but i need to it to complete the question. basically i've been set the question factorise $54-15x-25x^2$ . i know that you have to start by multiplying -25 and 54 which gives -1350. i then have to find two numbers that multiply together to give -1350 but also add together to give -15. is there a method i should use for finding the numbers i'm looking for?

thanks

**mathaddict** · August 26th 2009, 08:00 AM

Originally Posted by mark

i have a question to do with quadratic equations but what i'm asking isn't really about quadratics but i need to it to complete the question. basically i've been set the question factorise $54-15x-25x^2$ . i know that you have to start by multiplying -25 and 54 which gives -1350. i then have to find two numbers that multiply together to give -1350 but also add together to give -15. is there a method i should use for finding the numbers i'm looking for?

thanks

Let x and y be the 2 numbers ..

xy=-1350
x+y=-15

Then solve the simultaneous equation .

**mark** · August 26th 2009, 08:07 AM

can you show me how to solve that simultaneous equation please?

**mathaddict** · August 26th 2009, 08:17 AM

Originally Posted by mark

can you show me how to solve that simultaneous equation please?

Ok you have 2 equations here .

$xy=-1350$ --- 1

$x+y=-15$ --- 2

From 2 , make x the subject , then we get $x=-15-y$ ---3

Now we substitute 3 into 1 , we get

$(-15-y)(y)=-1350$

$ -15y-y^2+1350=0 $

Solve for y , i got $y=-45 , 30$

When $y=-45$ , $x-45=-15$ , $x=30$

and when $y=30$ , $x+30=-15$ , $x=-45$

So there are 2 sets of values for x and y which are -45 , 30 and 30 , -45 which are the same .

So when u say x is 30 , then y will be -45 and vice versa

Hope this helps .

**mark** · August 26th 2009, 08:33 AM

thanks,

i understood that up to $-15y - y^2 + 1350 = 0$ then you said "solve for y". how do you solve for y from that?

Mark

**mathaddict** · August 26th 2009, 08:36 AM

Originally Posted by mark

thanks,

i understood that up to $-15y - y^2 + 1350 = 0$ then you said "solve for y". how do you solve for y from that?

Mark

Make it positive so that it's easier to solve ..

$y^2+15y-1350=0$
$ (y+45)(y-30)=0 $

y=-45 , y=30

**aidan** · August 26th 2009, 09:27 AM

Originally Posted by mark

i have a question to do with quadratic equations but what i'm asking isn't really about quadratics but i need to it to complete the question. basically i've been set the question factorise $54-15x-25x^2$ . i know that you have to start by multiplying -25 and 54 which gives -1350. i then have to find two numbers that multiply together to give -1350 but also add together to give -15. is there a method i should use for finding the numbers i'm looking for?

thanks

mathaddict has provided the solution you sought.

Just curious, but could you post what you are going to do with those values?

**mark** · August 26th 2009, 10:19 AM

i would factorise them into $(54 - 45x) + (30x - 25x^2)$ which would then go to $9(6 - 5x) + x(30 - 25x)$ then $9(6 - 5x) + 5x(6 - 5x)$ then $(9 + 5x) (6 - 5x)$

**Soroban** · August 26th 2009, 10:47 AM

Hello, mark!

Factor: . $54-15x-25x^2$ .

I know that you have to start by multiplying -25 and 54 which gives -1350.
i then have to find two numbers with a product of -1350 and a sum of -15.
Is there a method i should use for finding the numbers i'm looking for?

I would factor out a -1: . $-\,(25x^2 + 15x - 54)$
. . and disregard the leading minus-sign for now.

Multiply the first and last coefficients: . $25\cdot54 \:=\:1350$

Note the sign of the last term of the quadratic: . $25x^2 + 15x - 54$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\uparrow$
. . If it is "+", we want a sum.
. . If it is "-", we want a difference.

We have "-", so we factor 1350 into two parts whose difference is the middle coefficient, 15.

How do we factor 1350 into two parts?
Divide 1350 by 1,2,3, . . . and keep the ones that "come out even".

. . $\begin{array}{ccc} \text{Factors} & \text{Difference} \\ \hline 1\cdot1350 & 1349 \\ 2\cdot675 & 5673 \\ 3\cdot450 & 447 \\ 5\cdot270 & 265 \\ 6\cdot226 & 219 \\ 9\cdot150 & 141 \\ 10\cdot135 & 125 \end{array}$
. . . $\begin{array}{cccccc}16\cdot90 &&&& 75 \\ 18\cdot75 &&&& 57 \\ 25\cdot54 &&&& 19 \\ 30\cdot45 &&&& 15 & \Leftarrow\:\text{ There!} \end{array}$

We want the middle term to be $+15x$ , so we will use $-30x \text{ and }+45x$

We have: . $25x^2 -30x + 45x - 54$

Factor: . $5x(5x-6) + 9(5x-6)$

Factor: . $(5x-6)(5x+9)$

Restore the leading minus-sign: . $-(5x-6)(5x+9)$

The answer can also be written: . $(6-5x)(9+5x)$

**mark** · August 26th 2009, 10:53 AM

thankyou soroban, that was actually very useful

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