Question

what is the zero of p(x)=(2x-1)(2x+1)​

Answer 1

p(x) = (2x - 1 )(2x + 1 ) = 0

• $\red{\implies}$ (2x)² - 1² = 0

• $\red{\implies}$ 4x² - 1 = 0

• $\red{\implies}$ 4x² = 1

• $\red{\implies}$ x² = $\frac{1}{4}$

• $\red{\implies}$ x = $\sqrt{\frac{1}{4}}$

• $\red{\implies}$ x = $±\frac{1}{2}$

Answer 2

Answer:

±1/2

Step-by-step explanation:

Here we can use the formula -

(a-b)(a+b) = a^2 - b^2

so, (2x-1)(2x+1) =

${(2x)}^{2} - {1}^{2} \\ = 4 {x}^{2} - 1$

Now to find its zero,

$4 {x}^{2} - 1 = 0 \\ = > 4 {x}^{2} = 1 \\ = > {x}^{2} = \frac{1}{4} \\ = > x = \sqrt{ \frac{1}{4} } \\ = > x = + \frac{1}{2} \: or \: - \frac{1}{2}$