Question

How do you solve `(2y + 3) ^ { 2} = ( y - 5) ^ { 2}`?

Answer 1

You can take the square root of both sides, BUT...

Explanation:

Remember there are two solutions to `x^2=a->x=+-sqrta`

So we have: `2y+3=+(y-5)and2y+3=-(y-5)`

(1)
`->2y+3=y-5->2y-y=-5-3->y=-8`

(2)
`->2y+3=-y+5->2y+y=5-3->3y=2->y=2/3`

Answer 2

See a solution process below:

Explanation:

First, square each side of the equation to eliminate the exponent:

`4y^2 + 12y + 9 = y^2 - 10y + 25`

Next, subtract `(color(red)(y^2) - color(red)(10y) + color(red)(25))` from each side of the equation to create a quadratic equal to `0`:

`4y^2 + 12y + 9 - (color(red)(y^2) - color(red)(10y) + color(red)(25)) = y^2 - 10y + 25 - (color(red)(y^2) - color(red)(10y) + color(red)(25))`

`4y^2 - color(red)(y^2) + 12y + color(red)(10y) + 9 - color(red)(25) = y^2 - color(red)(y^2) - 10y + color(red)(10y) + 25 - color(red)(25)`

`(4 - color(red)(1))y^2 + (12 + color(red)(10))y + (9 - color(red)(25)) = 0 +0 + 0`

`3y^2 + 22y - 16 = 0`

Then factor the left side of the equation:

`(3y - 2)(y + 8) = 0`

We can now solve each term on the left for `0` to find the solutions:

Solution 1

`3y - 2 = 0`

`3y - 2 + color(red)(2) = 0 + color(red)(2)`

`3y - 0 = 2`

`3y = 2`

`(3y)/color(red)(3) = 2/color(red)(3)`

`(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 2/3`

`y = 2/3`

Solution 2

`y + 8 = 0`

`y + 8 - color(red)(8) = 0 - color(red)(8)`

`y + 0 = -8`

`y = -8`

The Solutions Are: `y = 2/3` and `y = -8`