Question

If x-y=3,xy=10,find the value of x^2+y^2

Answer 1

Answer:

$\mathfrak{\large \red{\underline{\underline{Answer}}}} \\ \large \blue{x = 5 \: and \: x = - 2} \\ \large \blue{y = 2 \: and \: y = - 5} \\ \large \red{ \underline{ \underline{step = 1}}} \\ \large \green{eleminate \: the \: value \: of \: x \: in \: equation \: (2)} \\ \large \red{ \underline{ \underline{step =2}}} \\ \large \green{put \: the \: value \: of \: x \: in \: equation \: (1)} \\ \large \red{ \underline{ \underline{step =3}}} \\ \large \green{we \: can \: evaluate \: the \: value \: of \: y \: in \: equation \: (1)} \\ \large \red{ \underline{ \underline{step = 4}}} \\ \large \green{put \: the \: value \: of \: y \: in \: equation \: (3)} \\ \large \red{ \underline{ \underline{step =5}}} \\ \large \green{we \: can \: find \: the \: value \: of \: x}$

Step-by-step explanatio ${\large \red{ \underline{ \underline{to \: find = {x}^{2} + {y}^{2} }}}} \\ \large \green{x - y = 3.........(1)} \\ \large \green{xy = 10...........(2)} \\ \\ \large \green{from \: equation \: (2) \: we \: get} \\ \large \green{x = \frac{10}{y}...........(3) } \\ \large \green{put \: the \: value \: of \: x \: in \: equation \: (1)} \\ \large \green{ \frac{10}{y} - y = 3 } \\ \large \green{10 - {y}^{2} = 3y } \\ \large \green{ {y}^{2} + 3y - 10 = 0 } \\ \large \green{ {y}^{2} + 5y - 2y - 10 = 0 } \\ \large \green{y(y + 5) - 2(y + 5) = 0} \\ \large \green{(y - 2)(y + 5) = 0} \\ \large \green{y = 2 \: and \: y = - 5} \\ \large \blue{put \: y = 2 \: in \: equation \: (3)} \\ \large \pink{x = \frac{10}{2} = 5 } \\ \large \blue{put \: y = - 5 \: in \: equation \: (3)} \\ \large \pink{x = \frac{10}{ - 5} = - 2 }$

Answer 2

QUESTION:

If x-y=3,xy=10,find the value of x^2+y^2

TO FIND :

$value \: of \: {x}^{2} + {y}^{2}$

ANSWER:

Let;

$x - y = 3.....(eq.1)$

$xy = 10.....(eq.2)$

From equation 2 we get;

$xy = 10 \\ x = \frac{10}{y}$

putting the value of x in equation 1;

$\frac{10}{y} - y = 3$

$\red {(taking \: lcm)}$

$\frac{10 - {y}^{2} }{y} = 3y$

$\red {(cross \: multiplication)}$

${y}^{2} + 3y - 10 = 0$

now it is a quadratic equation we have to factorised;

Splitting the middle term.

In such a way that it's sum equal to +3 and product equal to -10.

${y}^{2} + 5y - 2y - 10 = 0 \\ y(y + 5) - 2(y + 5) = 0 \\ (y - 2)(y + 5) = 0$

so,

$\huge\pink {(y - 2)(y + 5) = 0}$

hence;

$y - 2 = 0 \\ y = 2 \\ \\ \\ or \\ \\ \\ y + 5 = 0 \\ y = - 5$

$\huge\blue {y = 2 \: and \: - 5}$

putting the value of y in equation 1 ;

When you = 2.

$x - y = 3 \\ x - 2 = 3 \\ x = 5$

$\huge\green {x = 5}$

when you = -5

$x - y = 3 \\ x - ( - 5) = 3 \\ x + 5 = 3 \\ x = - 2$

$\huge\orange {x = - 2}$

now;

When

x =5 and y = 2

then :

= x^2 + y^2

= 5^2 + 2^2

= 25 + 4

= 29

when;

x = -2 and Y = -5

then :

x^2 + y^2

= (-2)^2 + (-5)^2

= 4 + 25

= 29