Question

The sum of the squares of two numbers is 13 and their product is 6. Find (i) the sum of the two numbers. (ii) the difference between them

Answer 1

let the numbers be x and y

x² + y² =  13              xy = 6

(i)    (x+ y)² = x² + y² + 2xy

(x + y)²= 13 + 12                             (x + y)²=25

x + y = √25

x + y = 5

(ii)    (x - y)² = x² + y² - 2xy

(x - y)²= 13 - 12                             (x - y)²=1

REGARDS

x - y = √1

x - y =  1

Answer 2

Given,

The sum of the squares of two numbers is 13 and their product is 6.

To Find,

(i) The sum of the two numbers (ii) The difference between them

Solution,

We can solve the question as follows:

It is given that the sum of the squares of two numbers is 13 and their product is 6.

Let the first number be x and the other number be equal to y. Then,

$x^{2} + y^{2} = 13$

$xy = 6$

Now, we have to find:

(i) the sum of the two numbers

We know the identity,

$(x + y)^{2} = x^{2} + y^{2} + 2xy$

Substituting the given values in the above formula,

$(x + y)^{2} = 13 + 2*6$

$(x + y)^{2} = 13 + 12$

$(x + y)^{2} = 25$

Taking the square root on both sides,

$x + y = \sqrt{25} = 5$

The sum of the two numbers is 5.

(ii) the difference between them

We know the identity,

$(x - y)^{2} = x^{2} + y^{2} - 2xy$

Substituting the given values in the above formula,

$(x - y)^{2} = 13 - 2*6$

$(x - y)^{2} = 13 - 12$

$(x - y)^{2} =1$

Taking the square root on both sides,

$x - y = \sqrt{1} = 1$

The difference between the two numbers is 1.

Hence, the sum of the two numbers is 5 and the difference between them is 1.