Question

How do you solve `-3( x - 2) ^ { 2} - 33= 0`?

Answer 1

Solution : `x ~~ 2 + 3.32 i , x ~~ 2 - 3.32 i`

Explanation:

`-3 (x-2)^2 -33=0 or 3(x-2)^2= -33` or

`(x-2)^2= -11 or (x-2)= +-sqrt (-11)` or

`(x-2)=+-sqrt (11i^2)or x = 2 +- sqrt11 i`

`[i^2=-1] :. x = 2+sqrt 11 i , x = 2- sqrt 11 i` or

`x ~~ 2 + 3.32 i (2dp) , x ~~ 2 - 3.32 i (2dp)`

Solution : `x ~~ 2 + 3.32 i , x ~~ 2 - 3.32 i` [Ans]

Answer 2

See below.

Explanation:

`-3(x-2)^2-33=0`

Add `33` to both sides:

`-3(x-2)^2=33`

Divide both sides by `-3`:

`(x-2)^2=-11`

Take square roots of both sides:

`x-2 = sqrt(-11`

`x = 2+-sqrt(-11)`

Notice `sqrt(-11) => sqrt(11xx-1)=>sqrt(11)sqrt(-1)`

`sqrt(-1) =color(red)( i) => sqrt(11)color(red)(i)`

`color(red)(i)` is known as the imaginary unit.

This is sometimes written `isqrt(11)` to avoid any confusion.

So now we have:

`x= 2+-sqrt(11)color(red)(i)=> x= 2+-color(red)(i)sqrt(11)`

`x=2+color(red)(i)sqrt(11) , x= 2 - color(red)(i)sqrt(11)`

These are referred to as imaginary roots. These occur because the parabola neither crosses nor turns at the x axis, which indicates that no real number can satisfy the equation `-3(x-2)^2-33=0`

Graph: graph{y=-3(x-2)^2-33 [-30, 30, -400, 50]}