Math, asked by sirich1515, 7 hours ago

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

Answers

Answered by Athul4152
0

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} \\
Answered by abhinavjoshi88
0

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}

Similar questions