Quadratic

7) Let the number is read as "xy" where x and y are numbers from 0 to 9.
So from the given we have:

10x + y = 4(x + y)
xy = 18
So from the above conditions, we have:
10x + (18/x) = 4 [x + (18/x)]
10x2 + 18 = 4x2+ 72
6x2 = 54
x2 = 9
x = 3 or -3 (rejected)
Hence y = 6 and the two-digit number is 36.

8) Let the number is read as "xy" where x and y are numbers from 0 to 9.
So from the given we have:

10x + y = xy + 16
10y + x = 10x + y + 18
So from the above equations, we have:
9y = 9x + 18
y = x + 2
Sub this into (1):
10x + (x + 2) = x(x + 2) + 16
11x + 2 = x2 + 2x + 16
x2 - 9x + 14 = 0
(x - 7)(x - 2) = 0
x = 7 or x = 2
y = 9 or y = 4
Hence the two-digit number is 24 or 79.

9a) From the given, we can see that each side of the rhombus is 5cm in length.
Now, let the lengths of the rhombus are x and y with x being the longer one, i.e. x = y + 2.
Hence, by the Pyth. theorem, we have:
(x/2)2 + (y/2)2 = 52 = 25
[(y/2) + 1]2 + (y/2)2 = 25
(y2/4) + y + 1 + (y2/4) = 25
(y2/2) + y - 24 = 0
y2 + 2y - 48 = 0
(y - 6)(y + 8) = 0
y = 6 or y = -8 (rejected)
Therefore the diagonals of the rhombus are of lengths 6cm and 8cm.
b) The rhombus can be divided into 4 right-angled triangles, each with dimension 3cm by 4 cm.
So the area of the rhombus is given by: 4 x (1/2) x 3 x 4 = 24 cm2

(7)
Let the two-digit number be 10x+y.
10x+y=4(x+y)---(1)
xy=18---(2)
From (1):
10x+y=4x+4y
y=2x---(3)
Sub (3) into (2):
(2x)(x)=18
2x^2=18
x^2=9
x=3 or -3 (rejected)
x=3---(4)
Sub (4) into (3):
3y=18
y=6
Therefore the number is 36.

(8)
Let the number be 10x+y.
10x+y=16xy---(1)
(10y+x)-(10x+y)=18---(2)
From (2),
9y-9x=18
y=x+2---(3)
Sub (3) into (1):
10x+(x+2)=16x(x+2)
11x+2=16x^2+32x
16x^2+21x-2=0
x=[-21士(21^2-4(16)(-2))^(1/2)]/32
=[-21士(569)^(1/2)]/32
Which are not integers.
Therefore there is no solution.

(9)
(a)
Let the length of the shorter diagonal be x cm, then the longer one is (x+2) cm.
4{(x/2)^2+[(x+2)/2]^2}=20---(1)
(x^2)/4+[(x+2)^2]/4=5
x^2+x^2+4x+4=20
2x^2+4x-16=0
x^2+2x-8=0
(x-2)(x+4)=0
x=2 or x=-4 (rejected)
Therefore the length of the shorter diagonal is 2 cm, the longer one is 2+2=4 cm.
(b)
The area of the rhombus
=4{(x/2)[(x+2)/2]/2}
=4{(2/2)[(2+2)/2]/2}
=4[(1)(1)]
=4
Therefore the area of the rhombus is 4 cm sq.

7. xy=18
10x+y=4(x+y)
6x=3y
2x=y
=> 2x^2=18 =>x=3, y=6

8. 10x+y=16+(x)(y)
10y+x=10x+y+18 => 9y=9x+18 => y=x+2
10x+(x+2)=16+(x)(x+2) => 11x+2=16+x^2+2x => x^2-9x+14=0
solving the equatopn : (x-2)(x-7)=0 => x=2 and y=x+2=4 OR x=7 and y=9

9(a). x+x+y+y=20
x-y=2 => x=2+y
solve: (2+y)+(2+y)+y+y=20 =>4+4y=20 => y=4, x=2+y=6

9(b). area = (x)(y) = 4*6=24