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### Course: Algebra (all content)>Unit 12

Lesson 1: Solving square-root equations

# Solving square-root equations: no solution

Sal solves the equation √(3x-7)+√(2x-1)=0, only to find out that the single solution is extraneous, which means the equation has no solution.

## Want to join the conversation?

• What's wrong with leaving the terms on one side and squaring everything:
(sqrt(3x-7) + sqrt(2x-1))^2 = 0^2
• The radicals will not go away. When you square the left side, you need to use FOIL / distribution to do it. You would get: 3x - 7 + 2*sqrt(3x-7)*sqrt(2x-1) + 2x - 1
Notice, you now have a very complicated radical in the middle.
If you move one square root to the other side, then square the equation, both radicals will get eliminated and you end up with: 3x-7 = 2x-1
• Doesn't this equation have any possible solutions?
Not even a complex solution or something else of the sort?
Can that actually happen that a equation has no possible solution in any number system?
• Hi Rutwik, yes an equation can have no possible solution. It helps to think of the equation sqrt(3x-7) = - sqrt(2x-1) as two line graphs, one with the equation y = sqrt(3x-7) and the other y = -sqrt(2x-1) , with both graphs having no common intersection point.

Also, according to the Fundamental Theorem of Algebra, the equation { 3x - 7 = 2x + 1 } has only one root, which only exists if both sides have the same sign. In the initial equation, both sides do not have equivalent signs {sqrt(3x-7) = - sqrt(2x-1)} , which also hints that they do not have a root. Similarly, the equation {-sqrt(3x-7) = sqrt(2x-1)} has no root, while the equations {-sqrt(3x-7) = -sqrt(2x-1)} and {sqrt(3x-7) = sqrt(2x-1)} each have one root.
• At @ which is just when we isolate √(3x-7) on the left by subtracting √(2x-1) from both sides of the equation, to have
√(3x-7 = ` - `√(2x-1)
Wouldn't we know right then that there is no solution?
Under what conditions could we have the square root of one expression equal the negative of the square root of the other.
• It would still be possible for there to be a solution if both 3x-7 and 2x-1 were equal to 0. Obviously that is NOT the case, but the negative sign on the one radical doesn't eliminate the possibility of a solution by itself.
• Is there a certain rule of : whether the equation has one solution, two solution, or no solution
• Thats exactly what I was wondering. Is there a simple way of knowing if it is 2, 3, or 0 solutions ?
• at didn't sal make a mistake?
because (-√2x-1)^2=(-(2x-1))^0.5*2=(-2x+1)
but sal wrote (-√2x-1)^2=2x-1
• Your question involves the right hand side of the equation: (-sqrt(2x-1))^2.

Notice that everything inside the first set of parentheses is squared.
We can break down -sqrt(2x-1) into:
-1 * sqrt(2x-1)
Squaring them:
(-1)^2 * sqrt(2x-1)^2
Which results to our final answer for the right hand side:
1 * 2x-1, otherwise 2x-1.

So nope! Not a mistake, hopefully that helps.
• why not squre the equation without moveing the 2x-1 to the right side?
• You could do that; however, you can't simply square the left side and then remove the square roots. The reasoning why is this:
(a+b)² = a²+2ab+b²
(Let a = sqrt(3x-7) and b = sqrt(2x-1))

This property — more specifically the 2ab — makes squaring the entire left side mathematically labor intensive.
but if you're up w a challenge go for it:) — however, if this type of question is in a quiz, Sal's method is more faster.
• How come he did not just square both sides from the beginning?
• If he had squared both sides at the beginning, the radicals would not be eliminated.
To square `√(3x-7)+√(2x-1)` you would need to multiply:
(√(3x-7)+√(2x-1)) (√(3x-7)+√(2x-1)) using FOIL or extended distribution. You would end up with: `3x - 7 + 2√[(3x-7)√(2x-1)] + 2x -1`
Hope fully you can see that we still have a square root and the problem just got a lot more complicated than Sal's technique.
Hope this helps.
• Ok, so there must have been a way to solve it. My question is that could you have used FOIL and squared both sides of the equation shown above to get the answer?
• It would still be a no solution, but the procces will be much more complicated, because you will have a radical composed of both square roots in the middle.
(1 vote)
• It can also be done by using the (a+b)(a-b) = a^2-b^2 formula. You do that then by multiplying (√3x-7-√2x-1) in both sides.
1. (√3x-7+√2x-1) (√3x-7-√2x-1)=0 (√3x-7-√2x-1)
2. (√3x-7)^2 - (√2x-1)^2 = 0
3. (3x-7) - (2x-1) = 0
4. 3x-7-2x+1= 0
5. 3x-2x= 7-1
6. x= 6

Just laying it out there in case if someone thinks that is a better method