How do you solve `\sqrt { 5v - 9} = v - 1`?

Eliminate the radical by squaring both sides of the equation

`5v -9 = v^2 - 2v + 1`

Combine like terms:

`v^2-7v+10 = 0`

Solve the resulting quadratic:

`(v - 2)(v - 5) = 0`

`v = 2 and v = 5`

Check both answers:

`sqrt(5(2)-9) = 2-1`

`sqrt(1) = 1`

`1 = 1`

`sqrt(5(5)-9) = 5-1`

`sqrt(16) = 4`

`4 = 4`

Both answers check.

`sqrt(5v-9)=v-1` This type of equation is called 'Radical equation'.

Step 1) Square both sides of the equation to eliminate square root.
`[sqrt(5v-9)]^2=(v-1)^2`

Step 2) Use FOIL and simplify
`|5v-9|=v^2-2v+1`

Step 3) Consider when `|5v-9|` is positive or negative, then simplify.

i) `5v-9>0`, `v>9/5`

`5v-9=v^2-2v+1`
`−v^2+7v−10=0`
`v^2-7v+10=0`

Factor left side of the equation
`(v-5)(v-2)=0`
`v-5=0` or `v-2=0`

`∴v=2` or `v=5`
Check if it satisfies range `v>9/5` (Yes)
`∴v=2` or `v=5(∵v>9/5)` ------------- 1

ii) `5v-9<0`, `v<9/5`
`-(5v-9)=v^2-2v+1`
`v^2+3v+8=0`

Use quadratic formula
`v=-(3±sqrt41)/2`
Check if it satisfies range `v<9/5` (Yes)
But if you substitute `v=-(3±sqrt41)/2` to `sqrt(5v-9)=v-1`, It won't satisfy the equation.

`∴`No Solution `(∵`Doesn't satisfies when you substitute the value.`)` ------------- 2

Moderate 1 and 2,

`∴v=2` or `v=5`