Question

x+y=4,x^2+y^2=14 value of xand y?

Answer 1

x + y = 4 ................(i)
and,

x² + y² = 14

=> (x + y)² - 2xy = 14

=> 4² - 2xy = 14

=> 16 -2xy = 14

=> 16 - 14 = 2xy

=> 2 = 2xy

=> 2/2 = xy

=> y = 1/x .............(ii)

Using eqn(ii) in (i),we get

x + (1/x) = 4

=> x² + x = 4
=> x² + x - 4 =0
$= > x = \frac{ - 1 + \sqrt{ {1}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1}$
$= > x = \frac{ - 1 + \sqrt{1 + 16} }{2}$
$= > x = \frac{ - 1 + \sqrt{17} }{2}$
$= > x = \frac{ - 1 + \sqrt{17} }{2} \: \: \: \: or \: \: \frac{ - 1 - \sqrt{17} }{2}$
$if \: x = \frac{ - 1 + \sqrt{17} }{2}$
$y = \frac{2}{ - 1 + \sqrt{17} }$

Answer 2

Answer:

optionB 3+1=4 conditions apply but (3) ^2+(1) ^2=10

OptionC 2+√3+2-√3=4 and x^2+y^2 karne par bhi conditions apply hogi by option you can easly solve