Question

slove for x 4x^2-(a^2+b^2)x+a^2 b^2=0​

Answer 1

Answer:

Given,

4x2−2(a2+b2)x+a2b2=0

For general Quadratic Equation,

ax2+bx+c=0

a=4,b=−2(a2+b2),c=a2b2

According to formulae,

x=2a−b+b2−4ac=2a2

x=2a−b−b2−4ac=2b2

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Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, we have

4x² - 2(a² + b²)x + a²b² = 0

Therefore,

Constant term = a²b² = a² × b²

Coefficient of middle term = - 2(a² + b²)

then,

⇒ 4x² - 2(a² + b²)x + a²b² = 0

⇒ 4x² - 2a²x - 2b²x + a²b² = 0

⇒ (4x² - 2a²x) - (2b²x - a²b²) = 0

⇒ 2x(2x - a²) - b²(2x - a²) = 0

⇒ (2x - a²) (2x - b²) = 0

⇒ (2x - a²) = 0 or (2x - b²) = 0

⇒ x = a²/2 or x = b²/2

Hence, x = a²/2 or x = b²/2

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