Question

the sum of squares of 2 numbers is 41 and their product is 4. find the sum of two numbers​

Answer 1

Answer:

Sum of 2 numbers is 7.

Step-by-step explanation:

p^2+q^2 = 41

pq=4

Using the identity (a+b)^2

(p+q)^2 = p^2+q^2+2pq

            = 41+2(4)

           =  41+8

(p+q)^2=49 = 7^2

Removing square on both sides.

So ,

p+q=7

Answer 2

Given :

• sum of squares of 2 numbers = 41

• product = 4

To find :

• the sum of two numbers

Formula used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Solution :

Let the two numbers be x and y .

Sum of squares of 2 numbers = 41

product = 4

Therefore :-

${x}^{2} + {y}^{2} = 41$

$xy = 4$

We know that :-

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Therefore now :-

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

$\implies{(x+ y)}^{2} = 41 + 2(4)$

$\implies{(x+ y)}^{2} = 41 +8$

$\implies{(x+ y)}^{2} = 49$

$\implies{(x+ y)} = \sqrt{49}$

We know that 7² = 49

$\implies{(x+ y)} = \pm7$

Answer :

sum of two numbers = ±7

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$\large\bf\blue{Additional-Information}$

$\implies{(a+b)^2 = a^2 + b^2 + 2ab}$

$\implies{(a-b)^2 = a^2 + b^2 - 2ab}$

$\implies{(a+b)^3 = a^3 + b^3 + 3ab(a + b)}$

$\implies{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\implies{(a^3+b^3)= (a+b)(a^2 - ab + b^2)}$

$\implies{(a^3-b^3)= (a-b)(a^2 + ab + b^2)}$

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