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.
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
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.
To find the numbers when the sum of two numbers is given and twice their product is given, you can use a system of equations.
Let's assume the two numbers are x and y.
According to the given information, the sum of the two numbers is 9. So, we can write the equation as:
x + y = 9 ------(1)
Additionally, it is given that twice the product of the numbers is 40. This can be written as:
2xy = 40 ------(2)
Now, we have a system of equations. To solve them, we can use the method of substitution or elimination.
Let's use the substitution method.
From equation (1), we can express x in terms of y:
x = 9 - y ------(3)
Substituting the value of x from equation (3) into equation (2):
2(9 - y)y = 40
Expanding and rearranging the equation gives:
18y - 2y^2 = 40
Rearranging the equation gives a quadratic equation:
2y^2 - 18y + 40 = 0
Solving this equation will give you the values of y. Once you have the value of y, substitute it back into equation (1) to find the value of x.
For the given second problem, let one number be x and the other number be x + 4 (exceeding by 4).
According to the given information, the sum of their squares is 208. This can be written as:
x^2 + (x + 4)^2 = 208
Expanding and rearranging the equation gives:
x^2 + x^2 + 8x + 16 = 208
2x^2 + 8x + 16 = 208
Rearranging the equation gives a quadratic equation:
2x^2 + 8x - 192 = 0
Solving this equation will give you the values of x. Once you have the value of x, substitute it back into x + 4 to find the other number.
For the given third problem, let the length of the rectangular field be L and the width be W.
According to the given information, the area of the rectangular field is 108 square units. This can be written as:
L * W = 108 ------(1)
Additionally, it is given that the perimeter of the rectangular field is 42. This can be written as:
2L + 2W = 42 ------(2)
Now, we have a system of equations. To solve them, we can use the method of substitution or elimination.
Let's use the substitution method.
From equation (2), we can express L in terms of W:
L = (42 - 2W)/2 ------(3)
Substituting the value of L from equation (3) into equation (1):
(42 - 2W)(W)/2 = 108
Expanding and rearranging the equation gives:
21W - W^2 = 108
Rearranging the equation gives a quadratic equation:
W^2 - 21W + 108 = 0
Solving this equation will give you the values of W. Once you have the value of W, substitute it back into equation (3) to find the value of L.