Question

How do you solve `(3x - 2) ^ { 2} + 54= 0`?

Answer 1

`x=(2+3sqrt6i)/3,` `(2-3sqrt6i)/3`

Refer to the explanation.

Explanation:

Solve:

`(3x-2)^2+54=0`

Expand `(3x-2)^2` using the FOIL method.
https://www.ipracticemath.com/learn/algebra/foil-method-of-binomial-multiplication

`9x^2-12x+4+54=0`

Simplify `4+54` to `58`.

`9x^2-12x+58=0` is a quadratic equation in standard form:

`ax^2+bx+c=0`,

where:

`a=9`, `b=-12`, `c=58`

Solve using the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug in the known values and solve for `x`.

`x=(-(-12)+-sqrt((-12)^2-4*9*58))/(2*9)`

`x=(12+-sqrt(-1994))/18`

Prime factorize `sqrt(-1944)`.

`x=(12+-sqrt(-2*2^2*3*3^2*3^2))/18`

Apply rule: `sqrt(a^2)=a`

`x=(12+-2xx3xx3sqrt(-2xx3))/18`

`x=(12+-18sqrt(-6))/18`

Simplify `sqrt(-6)` to `sqrt6i`

`x=(12+-18sqrt6i)/18`

`x=(12+18sqrt6i)/18,` `(12-18sqrt6i)/18`

Simplify by dividing the integers by the GCF `6`.

`x=(2+3sqrt6i)/3,` `(2-3sqrt6i)/3`

Answer 2

`(2+-3isqrt6)/3`.

Explanation:

Given that, `(3x-2)^2+54=0 rArr (3x-2)^2=-54`.

In other words, `(3x-2)^2 lt 0`, so, no real solution.

To find soln. in `CC`, we have, `(3x-2)=sqrt(-54)=+-3isqrt6`.

`:. 3x=2+-3isqrt6`.

`:. x=(2+-3isqrt6)/3`.