X^2-16=0 solve for all real values of x

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asked Oct 21, 2022 179k views

1 Answer

Pankaj Shukla · Answer 1 · 2022-10-25T04:50:37+0000

Hi there! Use the difference of two squares formula below:

$\large \boxed{ {x}^(2) - {y}^(2) = (x + y)(x - y)}$

You need to know that what number times itself and get 16. 16 comes from 4×4 or 4².

Then,

$\large{ {x}^(2) - 16 = (x + 4)(x - 4)}$

$\large{(x + 4)(x - 4) = 0}$

Solve the equation for all real values of x.

$\large{x = 4, - 4}$

If you don't like using a formula. You can do this:

$\large{ {x}^(2) - 16 = 0} \\ \large{ {x}^(2) = 16} \\ \large{x = \pm √(16) } \\ \large{x = \pm 4}$

If you remember the square root well, the square root of 16 is 4×4. Pull out the two 4's and thus the square root of 16 is 4.

For the square both sides method above (second method), we can define that:

$\large \boxed{ \large{ {x}^(2) = a \longrightarrow x = \pm a }}$

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