How do you solve #sqrt(x+4) +sqrt(2x-1)=3sqrt(x-1)#?

1 Answer
Jun 6, 2016

Answer:

Explanation:

Before solving, let's determine the restrictions on the variable. A radical is undefined it the number inside is less than 0.

We must therefore set up an inequality and solve:

#sqrt(x + 4) >= 0#

#sqrt(2x - 1) >= 0#

and

#3sqrt(x - 1) >= 0#

Solving each we get that

#x >= -4#

#x >= 1/2#

and

#x >= 1#

We must pick the largest of these, since if we pick one of the smaller ones the equation will become undefined. The largest is #x >=1#; this is our restriction statement. To justify the point I made earlier in this paragraph, try substituting a number smaller than #1# into the equation. For example, #x= -2#

#sqrt(-2 + 4) + sqrt(2 xx -2 - 1) = 3sqrt(-2 - 1)#

#sqrt(2) + sqrt(-5) = 3sqrt(-3)#

In this case, two radicals are undefined. Now that we have defined any restrictions on the variable, we can proceed with solving. This can be started by squaring both sides

#(sqrt(x + 4) + sqrt(2x - 1))^2 = (3sqrt(x - 1))^2#

#x + 4 + 2sqrt(x + 4)sqrt(2x - 1) + 2x - 1 = 9(x - 1)#

#3x + 3 + 2sqrt(x + 4)sqrt(2x - 1) = 9x - 9#

# 2sqrt(x + 4)sqrt(2x - 1) = 6x - 12#

#sqrt(x + 4)(sqrt(2x - 1)) =( cancel(2)(3x - 6))/cancel(2)#

Since the radicals are both square roots, we can combine them in multiplication.

#sqrt((x + 4)(2x - 1)) = 3x - 6#

#sqrt(2x^2 + 8x - x - 4) = 3x - 6#

#(sqrt(2x^2 + 7x - 4))^2 = (3x - 6)^2#

#2x^2 + 7x - 4 = 9x^2 - 36x + 36#

#0 = 7x^2 - 43x + 40#

#0 = 7x^2 - 8x - 35x + 40#

#0 = 7x(x - 8/7) - 35(x - 8/7)#

#0 = (7x - 35)(x - 8/7)#

#x = 5 and 8/7#

Checking both answers in the original equation, we find that only #x = 5# works.

Therefore, our solution set is #{x = 5, x >= 1}#

Hopefully this helps!