(I)
Given equation is x²-8x = -16
⇛ x²-8x+16 = 0
⇛ x²-4x-4x+16 = 0
⇛ x(x-4)-4(x-4) = 0
⇛ (x-4)(x-4) = 0
⇛ x-4 = 0 or x-4 = 0
⇛ x = 4 or x = 4
(ii)
Given equation is x²+2x+1 = 0
⇛ x²+x+x+1 = 0
⇛ x(x+1)+1(x+1) = 0
⇛ (x+1)(x+1) = 0
⇛ x+1 = 0 or x+1 = 0
⇛ x = -1 or x = -1
(iii)
Given equation is 4x²-24x-36 = 0
⇛ 4(x²-6x-9) = 0
⇛ x²-6x-9 = 0
⇛ x²-2(x)(3) = 9
On adding 3² both sides then
⇛ x²-2(3)x)+3² = 9+9
⇛ (x-3)² = 18
⇛ (x-3) = ±√18
⇛ x-3 = ±√(2×9)
⇛ x-3 = ±3√2
⇛ x = 3±3√2
⇛ x = 3+3√2 or 3-3√2
Hello: here is a solution
Let c and n represent the numbers of pencil and pen boxes, respectively
.. 3c +2n = 6.00
.. 2c +4n = 8.00
You can halve the second equation and subtract it from the first to get
.. (3c +2n) -(c +2n) = 6.00 -4.00
.. 2c = 2.00
.. c = 1.00
Then
.. 1.00 +2n = 4.00 . . . . . half the original second equation with c filled n
.. 2n = 3.00
.. n = 1.50
A box of pencils costs $1.00; a box of pens costs $1.50.
Remember that transformation between Cartesian and polar system are:
x=r*cos(α)
y=r*sin(α)
From this we can conclude that:
r=√(x^2 + y^2)
Using trigonometry transformations we can write:
r=sin(2α) = 2sin(α)cos(α)
Now we can multiply both sides with r^2:
r^3 = 2(r*sin(α))*(r*cos(α))
Now using some replacements we can write:
(x^2 + y^2)^(3/2) = 2*x*y
