How do you solve # 4x=sqrt(6-4x)# and find any extraneous solutions?

1 Answer
Apr 8, 2017

Answer:

You can detect any extraneous solutions, if you restrict the argument of the square root to be greater than or equal to zero, then you can square both sides of the equation.

Explanation:

Given: #4x=sqrt(6-4x)#

Restrict the argument of the square root to be greater than or equal to zero:

#4x=sqrt(6-4x); 6-4x>=0#

Simplify the restriction:

#4x=sqrt(6-4x); 6>=4x#

#4x=sqrt(6-4x); 3/2>=x#

#4x=sqrt(6-4x); x<=3/2#

Square both sides:

#16x^2= 6-4x; x<=3/2#

#8x^2+2x-3=0; x<=3/2#

Use the quadratic formula:

#x = (-b+-sqrt(b^2-4(a)(c)))/(2a)#

#x = (-2+-sqrt(2^2-4(8)(-3)))/(2(8)); x<=3/2#

#x = (-2+-sqrt(100))/16; x<=3/2#

#x = (-12)/16 and x = 8/16#

#x = -3/4 and x = 1/2#