Question

if x^2+y^2=13 and xy=6, find fhe value of x-y

Answer 1

Answer: +1 or -1

Step-by-step explanation:

(x-y)²= x²+ y²- 2xy

(x-y)²= 13 - 2. 6

(x-y)²= 1

(x-y)= +1 or -1

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore x-y=1}}$

${\bold{\therefore x-y=-1}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies {x}^{2} + {y}^{2} = 13 \\ \\ \implies xy = 6 \\ \\ \underline \bold{to \: find : } \\ \implies x - y = ?$

• According to given question :

$\bold{using \: some \: identity \: of \: algebra : } \\ \implies {x}^{2} + {y}^{2} = 13 \\ \\ \implies ({x + y})^{2} - 2xy = 13 \\ \\ \bold{ putting \: value \: of \: xy : } \\ \implies ({x + y)}^{2} - 2 \times 6 = 13 \\ \\ \implies {(x + y)}^{2} - 12 = 13 \\ \\ \implies {(x + y)}^{2} = 13 + 12 \\ \\ \implies x + y = \sqrt{25} \\ \\ \implies x + y = 5 - - - - - (1) \\ \\ \implies xy = 6 \\ \\ \bold{\implies x = \frac{6}{y} - - - - - (2) } \\ \\ \bold{putting \: value \: of \: x \: in \: (1)} \\ \implies \frac{6}{y} + y = 5 \\ \\ \implies 6 + {y}^{2} =5 y \\ \\ \implies {y}^{2} - 5y + 6 = 0 \\ \\ \implies {y}^{2} - 3y - 2y + 6 = 0 \\ \\ \implies y(y - 3) - 2(y - 3) = 0 \\ \\ \implies( y - 2)(y - 3) = 0 \\ \\ \bold{\implies y = 2 \: and \: 3} \\ \\ \bold{putting \: value \: of \: y = 2 \: in \: (2)} \\ \implies \bold{x = 3} \\ \\ \bold{putting \: value \: of \: y = 3 \: in \: (2)} \\ \bold{\implies x = 2} \\ \\ \bold{for \: finding \: value : } \\ \bold{\implies x - y = 3 - 2 = 1} \\ \\ \bold{\implies x - y = 2 - 3 = - 1}$