Question

The sum of two numbers is 9.twice their product gives 40.find the numbers.
One number exceeds another by 4.the sum of their square is 208. What are the numbers.
The area of a rectangular field is 108square .the perimeter is 42. Find the length.

Answer 1

1.
x+y = 9
2xy=40

x=9-y
2(9-y)y=40
y²-9y+20=0
y= 9/2 ±sqrt(81/4 – 20)
y₁=5, y₂ =4 => x₁= 4, x₂=5

2.
x-y=4
x²+y² =208

x=4+y,
(4+y)² + y² =208,
y²+4y-96=0
y= - 2±sqrt(4+96)
y₁=8, x₁= 12 ,
y₂ = -12 ( this root doesn’t meet the condition as x>y)

3.
xy=108
2(x+y) =42

=> x+y=21 => x=21-y
(21-y)y=108
y²-21y +108=0
y= (21/2)±sqrt(441/4 -108)
y₁=12 y₂ = 9
x₁= 9 x₂=12

Answer 2

1. x + y = 9.
2xy = 40, xy = 20, X = 20/y.

x + y = 9.
20/y + y = 9,
Multiply by Y:
20 + y^2 = 9y,
y^2 - 9y + 20 = 0,
(x-4)(x-5) = 0,
Y = 4, and 5.

x + y = 9.
Solution set: (x,y)=(4,5),(5,4).
So the numbers are 4, and 5.

2. 1st number = X.
2nd number = x+4.
x^2 + (x+4)^2 = 208.
x^2 + x^2+8x+16 = 208,
2x^2 + 8x + 16 = 208,
Divide by 2:
x^2 + 4x + 8 = 104,
x^2 + 4x - 96 = 0,
(x-8)(x+12) = 0,
X = 8, and -12.
Solution set: 8,and 12. 0r -8, and -12.