Question

If x - y = 6 and xy = 10 find the value of x^2 + y^2​

Answer 1

$\sf\huge\underline{Explanation :-}$

Given :

• (x - y) = 6
• xy = 10

To find :

The value of x² + y²

Solution :

The algebraic identity we will be using is => (x - y)² = x² + y² - 2xy

According to question,

=> (x - y)² = x² + y² - 2xy

[To find the value of x² + y², we'll first put the value of (x - y) and xy]

=> (6)² = x² + y² - 2 × 10

=> 36 = x² + y² - 20

=> 36 + 20 = x² + y²

.°. x² + y² = 56

Hence, the value of x² + y² is 56.

$\sf\huge\underline{Verification :-}$

Let's put the value of (x - y), xy and x² + y² in the formula (x - y)² = x² + y² - 2xy and see if whether the value of x² + y² satisfies the equation.

=> (x - y)² = x² + y² - 2xy

=> 6² = 56 - 2 × 10

=> 36 = 56 - 20

=> 36 = 36

.°. L.H.S = R.H.S

Hence, proved.

Answer 2

Step-by-step explanation:

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

•x-y=6

•xy=10⠀⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

⠀⠀⠀⠀•$\bold{ x^{2}+y^{2}  }$ 

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

Here,

⠀x-y=6

xy=10

we know that,

$\boxed{\underline{\overline{\mathcal\color{gold}{(x-y)^2=x^2+y^2-2xy}}}}$

so,

According to the question,

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

$\mapsto{\sf{(6)^2=x^2+y^2-2×10   }}$      

$\mapsto{\sf{36=x^2+y^2-20   }}$      

$\mapsto{\sf{x^2+y^2=36+20   }}$      

$\rightsquigarrow$ $\bold{x^2+y^2=56   }$

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

$\therefore{\pink{\sf{The~ value~ of~ x^2+y^2~ is~ 56. }}}$