Find two numbers whose sum is 20, and the sum of whose squares is 208.
Answers
Required Answer :
The two numbers are :-
- When the first number = 28, then the second number = - 8
- When the first number = 32, then the second number = - 12
Given :
- Sum of two numbers = 20
- Sum of the square of two numbers = 208
To find :
- The two numbers
Solution :
Let the two numbers be x and y.
- First number = x
- Second number = y
According to the first condition,
→ Sum of the two numbers = 20
→ First number + Second number = 20
→ x + y = 20 -------(1)
According to the second condition,
→ Sum of the square of two numbers = 208
→ (First number)² + (Second number)² = 208
→ x² + y² = 208 -------(2)
Taking equation (1) :
→ x + y = 20
→ x = 20 - y ----(3)
Substituting equation (3) in (2) :
→ x² + y² = 208
→ (20 - y)² + y² = 208
Using identity :
- (a - b)² = a² + b² - 2ab
→ (20)² + (y)² - 2(20)(y) + y² = 208
→ 400 + y² - 40y + y² = 208
→ 400 + 2y² - 40y = 208
→ 2y² - 40y + 400 - 208 = 0
→ 2y² - 40y + 192 = 0
→ Taking 2 common :
→ 2(y² + 20y + 96) = 0
→ y² + 20y + 96 = 0
→ A quadratic equation is formed, whose product is 96y².
→ y² + 12y + 8y = 96 = 0
→ y(y + 12) + 8(y + 12) = 0
→ (y + 8)(y + 12) = 0
→ y + 8 = 0 or y + 12 = 0
→ y = - 8 or y = - 12
Substituting the value of y in (1) :
→ x + y = 20
When y = - 8
→ x + (-8) = 20
→ x - 8 = 20
→ x = 20 + 8
→ x = 28
When y = - 12
→ x + (-12) = 20
→ x - 12 = 20
→ x = 20 + 12
→ x = 32
Therefore,
- The first number = 28, the second number = - 8
- The first number = 32, the second number = - 12
Answer:
x + y = 20
x = 20 - y
x² + y² = 208
(20 - y)² + y² = 208
400 + y² - 40y + y² = 208
2y² - 40y + 400 - 208 = 0
2y² - 40y + 192 = 0 (dividing by 2)
y² - 20y + 96 = 0
y² - 12y - 8y + 96 = 0
y(y - 12) - 8(y - 12) = 0
(y - 12) (y - 8) = 0
if y - 12 = 0
y = 12
if y - 8 = 0
y = 8
therefore, y = 12 or 8
