Respuesta :
Answer:
The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169
Step-by-step explanation:
Given : the difference of the squares of the numbers is 5 and the product of the numbers is 6.
We have to find the sum of the squares of two numbers whose difference and product is given using given identity,
[tex](x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2[/tex]
Since, given the difference of the squares of the numbers is 5 that is [tex](x^2-y^2)^2=5[/tex]
And the product of the numbers is 6 that is [tex]xy=6[/tex]
Using identity, we have,
[tex](x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2[/tex]
Substitute, we have,
[tex](x^2+y^2)^2=(5)^2+(2(6))^2[/tex]
Simplify, we have,
[tex](x^2+y^2)^2=25+144[/tex]
[tex](x^2+y^2)^2=169[/tex]
Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169
Answer:
(x²+y²)² = 169
Step-by-step explanation:
We have given that
The difference of the squares of the numbers is 5.
x²-y² = 5
The product of the numbers is 6.
xy = 6
We have to find the sum of the squares of two numbers.
x²+y² = ?
We have given following formula:
(x²+y²)² = (x²-y²)²+(2xy)²
(x²+y²)² = (5)²+4(6)²
(x²+y²)² = 25+4(36)
(x²+y²)² = 25+144
(x²+y²)² = 169 which is the answer.