Question

If 4p² +1=2p; Then What is the value of
$4p {}^{2} + \frac{1}{4p {}^{2} }$
​

Answer 1

Step-by-step explanation:

Given that 4p² + 1 is 2p. We need to find out the value of 4p² + 1/4p².

Now,

2p = 4p² + 1 and 1/2p = 1/(4p² + 1)

And their squares will be the the square of 2p and 1/2p.

(2p)² = (4p² + 1)²

4p² = 16p⁴ + 1 + 8p²

Used identity: (a + b)² = a² + b² + 2ab

(1/2p)² = 1/(4p² + 1)²

1/4p² = 1/(16p⁴ + 1 + 8p²)

Therefore,

4p² + 1/4p² = 16p⁴ + 1 + 8p² + 1/(16p⁴ + 1 + 8p²)

Let's do the easy method.

4p² + 1 = 2p

4p² = 2p - 1

1/4p² = 1/(2p - 1)

Therefore,

4p² + 1/4p² = 2p - 1 + 1/(2p - 1)

4p² + 1/4p² = [(2p - 1)(2p - 1) + 1]/(2p - 1)

4p² + 1/4p² = [2p(2p - 1) - 1(2p - 1) + 1]/(2p - 1)

4p² + 1/4p² = (4p² - 2p - 2p + 1 + 1)/(2p - 1)

4p² + 1/4p² = (4p² - 4p + 2)/(2p - 1)