Question

If x+y = 5 and xy = 6. Find the value of x^2 + y^2 and x-y. ​

Answer 1

Given :

• x+y = 5

• xy = 6

To find :

• x²+y²

• x-y

Solution :

we know :-

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

Now, to find (x²+y²) put :-

• x+y = 5

• xy = 6

$\implies {(5)}^{2} = {x}^{2} + {y}^{2} + (2 \times 6)$

$\implies 25 = {x}^{2} + {y}^{2} + 12$

$\implies 25 - 12= {x}^{2} + {y}^{2}$

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

Now we will find x-y.

we know that :-

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

Now, to find (x-y) put :-

• x²+y² = 13

• xy = 6

$\implies{(x - y)}^{2} = 13- (2 \times 6)$

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

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

$\implies{(x - y)} = \sqrt{1}$

$\implies{(x - y)}= \pm1$

Answer :

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

${(x - y)}= \pm1$

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

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

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

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

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

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

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