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#