Question 628095
The sum of twice a number and 1 is equal to 3 times the positive square root of the number. Find the number. Write an equation and solve.

Let the number be x

{{{2x+1 = 3sqrt(x)}}}

square both sides

{{{(2x+1)^2 = (3sqrt(x))^2}}}


{{{4x^2+4x+1= 9x}}}

{{{4x^2-5x+1=0}}}

{{{4x^2-4x-x+1=0}}}

{{{4x(x-1)+1(x-1)=0}}}

(x-1)(4x-1)=0

either x-1 = 0 or 4x-1 =0

if x-1 =0
x= 1
if 4x-1 =0 then
4x=1
x= 1/4

x= 1OR 1/4