SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 118 of 200

Marks: +1, -0

If x+y=4,

x^2+y^2

= 14 and x > y, Then the correct value of x and y is:

3, 1
$2 − √2, √3$
$2 +√ 3, 2 − √3$
$2 + √3, 2√ 3$

Solution:

Given :

x+y=4,

and

x^2+y^2

= 14 ------(i)
Squaring both sides, we get :
⇒

(x + y)^ 2 = (4)^ 2

⇒

x^ 2 + y^ 2 + 2xy = 16

⇒

14 + 2xy = 16

⇒

2xy = 16 − 14 = 2

⇒

xy = 1

⇒

y=1/x

Substituting it in equation (i),⇒

x +1 / x = 4

⇒

x ^2 − 4x + 1 = 0

⇒

x ={4± √{ (−4) ^2−4(1)(1)}}/2

⇒

x ={4± √12}/ 2

⇒

x ={4±2 √3}/ 2

⇒

x = 2 ± √3

∵

x > y ⇒ x = 2 +√3

and

y = 2 - √3

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