Question

x+y=9
xy= 14
x^2-y^2=?​

Answer 1

Answer:

$\pm45$

Step-by-step explanation:

$x + y = 9\\y = 9 - x$

$xy = 14\\x(9-x) = 14\\9x - x^2 = 14\\x^2 - 9x + 14 = 0$

$x^2 - 7x - 2x + 14 = 0 \:\:\:|\:\:\: splitting\:the\:middle\:term\\x(x-7) - 2(x-7) = 0\\(x-7)(x-2) = 0$

$x = 2, \:7\\\therefore y = 9 - 2 = 7, \: 9 - 7 = 2$

Finally,

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

$(2 + 7)(2-7) = 9\cdot-5 = -45\\or\\(7+2)(7-2) = 9\cdot5 = 45$

$\therefore x^2 - y^2 = \pm45$

Answer 2

Question:-

➡ If x + y = 9 and xy = 14, find x² - y²

Answer:-

➡ x² - y² = ±45

Solution:-

Given that,

➡ x + y = 9

Squaring both sides, we get,

➡ (x + y)² = 9²

➡ x² + y² + 2xy = 81

➡ x² + y² + 2 × 14 = 81

➡ x² + y² = 81 - 28

➡ x² + y² = 53

➡ x² + y² - 2xy = 53 - 2xy

➡ (x - y)² = 53 - 28

➡ (x - y)² = 25

➡ x - y = √25

➡ x - y = ±5

So,

x² - y²

= (x + y)(x - y)

= 9 × ±5

= ±45

Hence,

➡ x² - y² = ±45