Question

Solve the equation.

2x^2 = -128

Answer 1

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

`\qquad \tt \rightarrow \: 8i`

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`\large \tt Solution \: :`

`\qquad\displaystyle \tt \rightarrow \: 2 {x}^(2) = - 128`

`\qquad\displaystyle \tt \rightarrow \: {x}^(2) = - ( 128)/(2)`

`\qquad\displaystyle \tt \rightarrow \: {x}^(2) = - 64`

`\qquad\displaystyle \tt \rightarrow \: x = √( - 64)`

`\qquad\displaystyle \tt \rightarrow \: x = √(( - 1) \sdot(64))`

`\qquad\displaystyle \tt \rightarrow \: x = √( 64) \sdot √( - 1)`

`\qquad\displaystyle \tt \rightarrow \: x = 8 \sdot i`

`\qquad\displaystyle \tt \rightarrow \: x = 8i`

[ i = iota, and i² = -1 ; (complex number) ]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer 2

`x=8i`

Explanation:

Given equation:

`2x^2=-128`

Divide both sides of the equation by 2:

`\implies (2x^2)/(2)=(-128)/(2)`

`\implies x^2=-64`

Square root both sides:

`\implies √(x^2)=√(-64)`

`\implies x^2=√(-64)`

Rewrite -64 as 8² · -1:

`\implies x=√(8^2 \cdot-1)`

`\textsf{Apply radical rule} \quad √(ab)=√(a)√(b):`

`\implies x=√(8^2)√(-1)`

`\textsf{Apply radical rule} \quad √(a^2)=a, \quad a \geq 0:`

`\implies x=8√(-1)`

`\textsf{Since $√(-1)=i$}`

`\implies x=8i`

Imaginary numbers

Since there is no real number that squares to produce -1, the number
`√(-1)` is called an imaginary number, and is represented using the letter i.

An imaginary number is written in the form
`bi`, where
`b \in \mathbb{R}`.