4x2-4ax+(a2-b2)=0 solve the equation using factorization method .


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  • 0

Solution :-

Here, we have

4x² - 4ax + (a² - b²) = 0

Constant term = (a² - b²) = (a - b) (a + b)

Coefficient of middle term = - 4a

Coefficient of middle term = - 4a = - [2(a + b) + 2(a - b)]

Therefore,

⇒ 4x² - 4ax + (a² - b²) = 0

⇒ 4x² - [2(a + b)x - 2(a - b)]x + (a + b) (a - b) = 0

⇒ 4x² - 2(a + b)x - 2(a - b)x + (a + b) (a - b) = 0

⇒ [4x² - 2(a + b)x] - [2(a - b)x - (a + b) (a - b)] = 0

⇒ 2x[2x - (a + b)] - (a - b) [2x - (a - b)] = 0

⇒ [2x - (a + b)] [2x - (a - b)] = 0

⇒ [2x - (a + b)] = 0 or [2x - (a - b)] = 0

⇒ 2x = a + b or 2x = a - b

⇒ x = a + b/2, a - b/2

Hence, x = a + b/2, a - b/2

  • -2

Answer:

Step-by-step explanation:

Solution :-

Here, we have

4x² - 4ax + (a² - b²) = 0

Constant term = (a² - b²) = (a - b) (a + b)

Coefficient of middle term = - 4a

Coefficient of middle term = - 4a = - [2(a + b) + 2(a - b)]

Therefore,

⇒ 4x² - 4ax + (a² - b²) = 0

⇒ 4x² - [2(a + b)x - 2(a - b)]x + (a + b) (a - b) = 0

⇒ 4x² - 2(a + b)x - 2(a - b)x + (a + b) (a - b) = 0

⇒ [4x² - 2(a + b)x] - [2(a - b)x - (a + b) (a - b)] = 0

⇒ 2x[2x - (a + b)] - (a - b) [2x - (a - b)] = 0

⇒ [2x - (a + b)] [2x - (a - b)] = 0

⇒ [2x - (a + b)] = 0 or [2x - (a - b)] = 0

⇒ 2x = a + b or 2x = a - b

⇒ x = a + b/2, a - b/2

Hence, x = a + b/2, a - b/2

  • -2
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