How do you solve #x= sqrt(2x+3)#?

1 Answer
Aug 19, 2015

Answer:

#x = 3#

Explanation:

Right from the start, you know that you can only use positive values of #x#, since taking the square root of a real positive number will always produce a positive value.

This means that you need to have #x >=0#.

With this in mind, start by squaring both sides of the equation

#(sqrt(2x+3))^2 = x^2#

#2x + 3 = x^2#

Rearrange this equation by moving all the terms on one side

#x^2 - 2x - 3 = 0#

You can find the solutions to this quadratic equation by using the quadratic formula, which for a general form quadratic

#color(blue)(ax^2 + bx + c = 0)#

allows you to find the roots by using the formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#

In your case, you can write

#x_(1,2) = (-(-2) +- sqrt((-2)^2 - 4 * 1 * (-3)))/(2 * 1)#

#x_(1,2) = (2 +- sqrt(16))/2#

#x_(1,2) = (2 +- 4)/2 = {(x_1 = (2 + 4)/2 = 3), (x_2 = (2 - 4)/2 = -1) :}#

SInce #x_2 = -1# does not satisfy the condition #x>=0#, your original equation will have only one solution, #x = color(green)(3)#.

You can do a quick check to make sure that the calculations are correct

#sqrt(2 * 3 + 3) = 3#

#sqrt(9) = 3#

#3 = 3" "color(green)(sqrt())#