Question

Help w/ Quadratic Equations 5?

I know these are three together, please bare with me. Please solve the following. Show your work and be as visual as possible.

Answer 1

`Q 23.` Sides of the squares to be cut out will be `2cm`

`Q 24.` Width of pavement is `1.5m`

`Q 25." " x=5cm`

Explanation:

`Q 23:` The squares which are cut from each corner obviously have to have the same length sides, otherwise the sides of the box formed would have different heights.

Let the length of the squares be `x`

The thing to note is that you are cutting a square at EACH corner, so the length and breadth of the sheet of cardboard are both shorter by `2x`

There is a rectangle left in the middle.

New length `= 15-2x`
New breadth `= 10-2x`

Once you have cut out the four squares, the area of the central rectangle must be `66cm^2`

`l xxb=A" "larr` use this to make an equation.

`(15-2x)(10-2x) =66`

`150-30x-20x +4x^2 = 66" "larr` make equal to `0`

`4x^2 -50x +84 =0" "larr div 2`

`2x^2 -25x+42=0" "larr` factorise

`(2x-21)(x-2) =0`

Setting each factor equal to `0` and solving gives

`x = 21/2 = 10.5" "larr` reject as too big.
`x =2`

Sides of the squares to be cut out will be `2cm`

Check: `15-4 = 11 and 10-4 = 6`

`11 xx6 = 66cm^2`
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`Q 24.` The pool and the paving form a larger rectangle with

`l = (12+2x) and b = (6+2x)`

a) The area of the paving = area of rectangle - area of pool

`A = (12+2x)(6+2x) -12xx6`

`A=72+24x+12x+4x^2 -72`

`:. A= 4x^2 +36x`

b) The area of the paving equals `7/8` of area of pool.

`4x^2 +36x = 7/8 xx72= 63`

`:. 4x^2+36x-63 =0`

c) Solve for `x` by factorising:

`(2x-3)(2x+21)=0`

`2x-3=0 rarr x =1.5`

`2x+21 =0 rarr x = -10.5 " "`(reject as invalid negative)

`:.x =1.5` is the width of the paving.

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`Q25." " x=5cm`

The base of the magnet is the area between 2 circles. Mathematically this is called a 'washer',

Its area is the difference between the area of the larger outer and smaller inner circle. Let's call the two radii `R and r`

`Area = color(magenta)(piR^2-pir^2)`

In this case the two radii are `xcm and (x+2)cm`

The Volume of the magnet is `120pi` and is found from:

`color(magenta)("Area of the base") xx color(blue)("height")` (where the height is also `color(blue)(x)cm`)

Now we can put this all together:

`V= color(magenta)((pi(x+2)^2 -pir^2)) xxcolor(blue)(x) = 120pi`
`V= color(magenta)(pi((x+2)^2 -x^2)) xxcolor(blue)(x) = 120pi" "larr` factor out `pi`

`V= color(magenta)(cancelpi(x^2+4x+4 -x^2) xxcolor(blue)(x) = 120cancelpi" "divpi`

`x(4x+4) = 120`

`4x^2 +4x-120=0`

`x^2+x-30=0`

`(x-5)(x+6)=30`

`x=5 or x=-6"` (reject)
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