If x² + y² = 37 and xy = 6; find :

(i) x + y

(ii) x - y

(iii) x² - y²

(i) Given, x² + y² = 37 and xy = 6

We need to find the value of (x + y):

$⇒ (x + y)^2 = x^2 + y^2 + 2xy$

Substituting the value of x² + y² and xy,

$⇒ (x + y)^2 = 37 + 2 \times 6\\[1em] ⇒ (x + y)^2 = 37 + 12\\[1em] ⇒ (x + y)^2 = 49\\[1em] ⇒ x + y = \sqrt{49}\\[1em] ⇒ x + y = \pm 7$

Hence, x + y = $\pm$ 7.

(ii) We need to find the value of (x - y):

$⇒ (x - y)^2 = x^2 + y^2 - 2xy$

Substituting the value of x² + y² and xy,

$⇒ (x - y)^2 = 37 - 2 \times 6\\[1em] ⇒ (x - y)^2 = 37 - 12\\[1em] ⇒ (x - y)^2 = 25\\[1em] ⇒ x - y = \sqrt{25}\\[1em] ⇒ x - y = \pm 5$

Hence, x - y = $\pm$ 5.

(iii) $x^2 - y^2$ = (x - y)(x + y)

Using (i) and (ii),

When (x - y) = 7 and (x + y) = 5,

⇒ $x^2 - y^2$ = 7 x 5

= 35

When (x - y) = -7 and (x + y) = 5,

⇒ $x^2 - y^2$ = -7 x 5

= -35

When (x - y) = 7 and (x + y) = -5

⇒ $x^2 - y^2$ = 7 x (-5)

= -35

When (x - y) = -7 and (x + y) = -5

⇒ $x^2 - y^2$ = (-7) x (-5)

= 35

Hence, $x^2 - y^2= \pm$ 35.