Question

The difference between the squares of two numbers is 15. Three times the square of the first number increased by the square of the second number is 49. Find the numbers

Answer 1

1, 4 or -1, -4

Explanation:

Let the two numbers be x and such that x > y.

According to the first condition:

`x^2 -y^2= 15`

`\implies x^2 =y^2+15`......(1)

According to the second condition:

`3x^2 +y^2= 49`

`\implies 3(y^2+15) +y^2= 49`

(From equation 1)

`\implies 3y^2+45 +y^2= 49`

`\implies 4y^2 =49-45`

`\implies 4y^2 =4`

`\implies y^2 =(4)/(4)`

`\implies y^2 =1`

`\implies y =\pm 1`

When y = 1

`\implies x^2 =(1)^2+15=1+15=16`

`\implies x =\pm 4`

When y = -1

`\implies x^2 =(-1)^2+15=1+15=16`

`\implies x =\pm 4`

Thus, the required numbers are either 1, 4 or -1, -4

Answer 2

Answer: 256,1

Step-by-step explanation:

`√(x) - √(y) =15\\√(y) = 49-3√(x) \\\\√(x) -(49-3√(x) ) =15 \\so, x=256\\substitute \{x}\ :\\\\√(y) = 49-3√(256) =y=1`

so, 256=x

and, y=1

Hope this Helped!