show that (3x+7)^2-8=(3x-7)^2
Answers
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❄❄Here is your Answer
As per identity :- a²+b²+2ab, So,
It is proved that
➡(3x+7)²-84x= (3x-7)²
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Step-by-step explanation:
An identity is true only for certain values of its variables. An equation is not an identity.
The following are the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
(a – b)(a + b) = a² – b²
Another useful identity is
(x + a) (x + b) = x² + (a + b) x + ab
If the given expression is the difference of two squares we use the formula
a² –b² = (a+b)(a-b)
• The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.
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Solution:
1) (3x + 7)² – 84x = (3x – 7)²
LHS = (3x)²+2×3x×7+(7)²-84x
9x² + 42x + 49 - 84x
9x² + 42x - 84x+49
= 9x² - 42x + 49
RHS = (3x)²-2×3x×7+(7)²
9x² - 42x + 49
LHS = RHS
2) (9p – 5q)²+ 180pq = (9p + 5q)²
LHS =(9p – 5q)²+ 180pq
(9p)²-2×9p×5q+(5q)²
= 81p² - 90pq + 25q² + 180pq
= 81p² - 90pq + 180pq + 25q²
= 81p² + 90pq + 25q²
RHS = (9p + 5q)²
(9p)²+2×9p×5q+(5q)²
81p² + 90pq + 25q²
LHS = RHS
4) LHS (4pq + 3q)²– (4pq – 3q)²
(4pq)²+2×4pq×3q+(3q)² - ((4pq)²- 2×4pq×3q+(3q)²)
= 16p²q²+ 24pq²+ 9q² - 16p²q² + 24pq² - 9q²
= 48pq²
RHS= 48pq²
LHS = RHS
5) LHS= (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
a² - b² + b² - c² + c² - a²
= 0
RHS= 0
LHS===========================================================
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