Question

HELP.
Mr Brown asks a garden designer to make a square flower bed. he cost of paving the border of the flower bed is $20 per metre. The cost of the flowers and soil in the flower bed is $60 per m^2. If the total cost of making the flower bed is $3500, find the length of a side of the flower bed,

Answer 1

Draw a square and call a side of it x.

Then u have 2x worth of border and x^2 worth of bed and the sum of the two is 3,500.

Solve the equation u get.

Answer 2

:O omg thank u . i didn't realise that

Answer 3

um im a little stucked. could u help me solve?

Answer 4

OK, you have x^2 + 2x = 3500 which is a quadratic, you know the quadratic formula?

Answer 5

The one that goes - b plus minus sqrt(.......

Answer 6

If you don't know it, I will show u...

Answer 7

okay i get it :D my answer is 7m

Answer 8

Let's see 4x of border is 28*20 = 560

and 49*60 is 2940 for a total of 3500

looks good to me.

(Even though I misled you about the border, putting 2x instead of 4x:-)

Answer 9

:D haha. thanks

Answer 10

Did you do it with quadratic or just sort of guess it?

Answer 11

quadratic equation. :)
heres my working.
let 1 side of the square flower bed be x
cost of 4 sides=$80x
cost of area=$60x^2
60x^2+80x=3500
60x^2+80x-3500=0
(10x-70)(6x+50)=0
either 10x-70=0 or 6x+50=0
x=7 or x=-8 1/3 (rejected)
my answer is 7m

Answer 12

Perfect:-)

Answer 13

thanks for the medal :)

Answer 14

Deserved.

Answer 15

:) hey can i ask u smth?
when do we use quadratic to solve and when do we use simultaneous(elimination and substitution method)?

Answer 16

The essential difference is the number of variables (unknowns).

In the above we only had one, x.

If you have two unknowns, then you need two equations (and three for three,etc)

and then you would solve these by some method like elimination or substitution.

In fact what you are doing is using the info you have to reduce the two equations in 2 unknowns to a single equation in 1 unknown which you can solve (and then substitute back to get the other unknown).

Answer 17

okay :) so does that mean that we use simultaneous(substitution n elimination) for equations with more than 1 variable? & quadratic for equations with 1 variable?

Answer 18

Roughly speaking, yes (of course some equations in 1 variable do not involve a squared term like x +3 = 4, linear as opposed to quadratic, or may involve higher powers, like a cubic).

You will also learn some other methods when there is more than 1 variable.

Answer 19

o.O yeahh okay :) thanks and see u!

Answer 20

And you, ciao!