Question

Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2​

Answer 1

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The given problem can be solved using the expansions of the forms (a+b)², and (a-b)².

1. The given equation is  (11-p q + 4q)² - (11-p q - 4q)² = 176 p q².

2. Consider the formulas (a+b)², and (a-b)². They are formulated as,

(a+b)² = a² + b² + 2ab,

(a-b)² = a² + b² - 2ab.

3. Using the above formulae, expand the LHS of the given equation,

=> ( 121 p²q² + 16q² + 88pq² ) - ( 121 p²q² + 16q² - 88pq² ),

=> 121 p²q² + 16q² + 88pq² - 121 p²q² - 16q² + 88pq², ( solve the equations by subtracting like terms),

=> 0 + 0 + 176 pq²,

=> 176 pq² which is equal to the value of the RHS.

Hence, equality is proved.

Therefore, the  equation (11-p q + 4q)² - (11-p q - 4q)² = 176 pq² is correct and verified.

Answer 2

Answer:

The given problem can be solved using the expansions of the forms (a+b)², and (a-b)².

1. The given equation is (11-p q + 4q)² - (11-p q - 4q)² = 176 p q².

2. Consider the formulas (a+b)², and (a-b)². They are formulated as,

(a+b)² = a² + b² + 2ab,

(a-b)² = a² + b² - 2ab.

3. Using the above formulae, expand the LHS of the given equation,

=> ( 121 p²q² + 16q² + 88pq² ) - ( 121 p²q² + 16q² - 88pq² ),

=> 121 p²q² + 16q² + 88pq² - 121 p²q² - 16q² + 88pq², ( solve the equations by subtracting like terms),

=> 0 + 0 + 176 pq²,

=> 176 pq² which is equal to the value of the RHS.

Hence, equality is proved.

Therefore, the equation (11-p q + 4q)² - (11-p q - 4q)² = 176 pq² is correct and verified.