Question

2
2.19 If x +y =12 and xy =14, find the value of (x+​

Answer 1

Step-by-step explanation:

Given:-

x+y=12 , xy=14

To find :-

value of x+y)^2

Solution

$(x + y) = 12 \\ = > {(x +y )}^{2} = 12 \\ = > { x }^{2} + 2xy + {y}^{2} = 12 \\ = > {x}^{2} + {y}^{2} + 2xy= 12 \\ = > {x}^{2} + {y}^{2} + 2(14) = 12 \\ = > {x}^{2} + {y}^{2} + 28 = 12 \\ = > {x}^{2} + {y}^{2} = 12 - 28 \\ = - 16 \\ = > (x + y)(x + y) = - 16 \\ = > 12 \times 12 = - 16 \\ = > 144 = - 16 \\ = > x + y = \frac{144}{ - 16} \\ = - 9$