Question

if (a+b)(a+b)=289 &(a-b)(a-b)=121 find (a+b)(a-b)=?

Answer 1

(a+b)(a+b) = 289
(a+b)^2 = 289
(a+b)^2 = 17^2
(a+b) = 17

(a-b)(a-b) = 121
(a-b)^2 = 11^2
(a-b) = 11

(a+b)(a-b) = 17 × 11= 187
please mark as brainlist please please mark as brainlist

Answer 2

Hi friend
----------------
Your answer
---------------------
Given that => (a + b)(a + b) = 289

So,
-------

(a + b)(a + b) = 289

=> (a + b)² = 289

=> (a + b) = √289

=> (a + b) = 17

Again,
------------


(a - b)(a - b) = 121

=> (a - b)² = 121

=> (a - b) = √121

=> (a - b) = 11

So,
------

(a + b)(a - b) = 17 × 11

(a + b)(a - b) = 187

HOPE IT HELPS

#ARCHITECTSETHROLLINS

✯ BRAINLY STAR ✯