If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra 2>Unit 10

Lesson 3: Extraneous solutions

# Equation that has a specific extraneous solution

Sal finds the value of d for which √(3x+25)=d+2x has an extraneous solution at x=-3.

## Want to join the conversation?

• Wouldn't (d+2x)^2 result in (d^2+4xd+4x^2)? I keep trying it at getting the same result.
• You are correct.
If you think you have spotted an error, please report it.
• It's -sqrt(16) the same as sqrt(-16)?
• The square root of 16 times negative 1, would be: `-sq rt(16) = -(4) = -4`
The square root of -16 itself however, would be 4i.
So they aren't the same; one is -4, the other is 4i. Did that help?
• When asked to pause the vid and try ourselves, I just inputted -3 for x and solved for d, which came back as:
4=d-6 or 10=d
As long as d is not equal to 10 then x=-3 would be an extraneous solution. However, Sal got 2=d, so why did he go for something so specific? What am I missing here?
If someone would explain Sal's reasoning to me, I'd be immensely grateful.
• I did it the same way as you initially and after some consideration, I think the issue is that an extraneous solution doesn't simply mean a solution that makes the equation untrue; it is, by definition, "a solution that emerges from solving the problem but is not a valid solution to the problem". The method you used was to simply solve for a d that would prove the equation true, then concluded that any other d would not be true. However, for it to be an extraneous solution, we would have had to come across a value for d through the process of solving the equation that does not actually fit the original equation.

For example, in previous videos Sal demonstrates that there occurs extraneous solutions which solve an equation that was a valid algebraic manipulation of the original equation, but not the original equation itself due to using an operation that cannot be reversible ie taking the square root of both sides.

So in conclusion, an extraneous solution is not simply just any number plucked out that does not prove true the original equation, but a solution that was algebraically extracted through valid operations that does not solve the original equation due to the reversibility of operations used.
• Also, at Sal says, "And so when you solve this purple equation... a quadratic... you'll get two solutions. And it turns out, one of the solutions is going to be for this yellow equation, and one of the solutions is going to be for the purple [red alternate one on the left] equation."

But, in the lesson 'Solving square-root equations: two solutions', we saw a situation where we squared both sides of a radical equation, got a quadratic, ended up with two solutions, and neither solution was extraneous. There is still that alternate equation looming about that may bring about extraneous solutions, since we are squaring both sides of our radical equation, yet neither solution satisfies the alternative extraneous-spawning equation that we only see when we reverse, they both are valid and satisfy the original equation. So, how can he say here that one solution will be for the original equation and one will be for the alternate 'hidden' equation - it seems that this can't be guaranteed, as it's possible for neither one to be for the alternate equation.

Does he just know that it will be the case for this exact problem, but it's not necessarily true in general? Or is it something else about this equation specifically that I'm missing? Because clearly there are situations where one of the solutions is not for the alternate equation, and rather both are for the original... thanks to anyone who can explain :)
• Why did he write the negative equation (the one in pink on the left)?
• He was just showing that even if you had a negative square root, it would still work out to the same quadratic, however, the solutions to that quadratic may not work out for the negative square root and the positive one.
• did anyone else see that Sal made a mistake and said that the square root of (3x+25) = d+2x equals 3x+25=d^2+4dx+x^2 instead of 3x+25=d^2+4dx+4x^2?
• Yeah, but you will see that there is a box that says around , "Sal wrote x² but it should be 4x².", so next time check if there a correction box, if not report the problem (if you are in full screen then it shouldn't appear).
• How would squaring two sides keep each side equal? You would be doing different things to each side. For example sqrt(x+3)=5x then you would be multiplying one side by x+3 and the other by 5x
(1 vote)
• Remember, the 2 sides are currently equal, but shown using different symbols.
So, you are squaring equal values.
If you have: sqrt(49) = 7
These are currently equal. Both sides are 7, but written in different ways.
Now, square both sides: sqrt(49)^2 = 7^2
You get: 49 = 49
They are still equal.
Hope this helps.
• I don't understand why this concept exists. Aren't negatives still square roots of a number? If both -2^2 = 4 and 2^2 = 4, then why aren't they both square roots of 4? If -2 is a square root of 4, why does sqrt(4) not equal -2?
• What is the definition of Radical solution?