2 Answers

C Black · Answer 1 · 2024-01-05T14:32:36+0000

$\huge\bold \large \bf{Solution }$

$\rightarrow \large \displaystyle \sf \: {x}^(2) = 10 \:$

Transposing square on RHS , We get

$\rightarrow \large \displaystyle \sf \: x \: = √(10)$

We know that,

10 does not even have a square root , so,

$\rightarrow \large \displaystyle{\bf{ \pink{x \: = \pm \: √(10) }}}$

We know that the value of root is both negative and positive.

$\purple {\rule{190pt}{4pt}}$

Shnizlon · Answer 2 · 2024-01-08T20:38:21+0000

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Lets solve ~

$\qquad \sf  \dashrightarrow \: {x}^(2) = 10$

[ taking square root on both sides ]

$\qquad \sf  \dashrightarrow \: \sqrt{ {x}^(2) } = √(10)$

$\qquad \sf  \dashrightarrow \: x = \pm √(10)$

Both the values are appropriate here, as the square of both negative or positive value is same [positive value]

Please log in or register to add a comment.