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# Solving square-root equations: two solutions

Sal solves the equation 6+3w=√(2w+12)+2w which has two solutions.

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• Why are both solutions to this equation non-extraneous? I thought squaring a radical provides two answers--one for if the radical is positive, and one for if it is negative. Thus, any equation we solve by squaring the radical will provide us with two answers, one of which will always be extraneous. Why then do both of the solutions to this equation fit with the positive square root, while only one (-6) fits with the negative?
• Nicolas, hi!

If you think about that, there are two potential solutions to the equation aˆ2=bˆ2 (which is an equation you get when solving radical equations: you square both sides):
1) a=b (including the case a=0 and b=0)
2) a=-b (which is the same as -a=b)

If there is a solution for -a=b, it is going to be extraneous to the original equation a=b. However, in this case we have 2 different values of w that satisfy our a=b case: -4 and -6.

The two solutions for a=b case result from the fact that after squaring we get a quadratic equation with 2 solutions for the case a=b. However, it does not necessarily have to be 2 solutions, it can be one or more than 2, including extraneous. It all depends on which equation we get after squaring.

Hope this helps.

Have fun:)
• at how did 12w came in the equation
w.sq+12w+36
• ( 6 + w ) ^2 = ( 6 + w ) ( 6 + w )
Then use FOIL to multiply the two binomials:
First terms : 6*6 = 36
Inner terms : +6w
Outer terms : + 6w
Last terms : w*w = w^2
• Can it be determined in advance if an equation will have extraneous solutions, without actually having to plug in the answers in the original equation afterwards?
• Hello Hodorius,

This is technically true. You can determine which possible solutions would be extraneous if they were indeed solutions.

This sounds complicated, but it's really just a matter of saying, "if solution x is true, will it be extraneous?"

I won't explain how to go all the way through it, as it is somewhat difficult to explain, but it basically is identifying which possible solutions would make the denominator equal to zero.

As for identifying whether an expression will have one of it's solutions guaranteed to be extraneous, I do not know of one.

For now, the most efficient way is to just do as Sal did above; solve the equation and check them against your original equations.

Hope This Helped!
• What property allows Sal to go from a quadratic equation of (w+4)(w+6) at ; and simply isolate each part of the equation, (w+4=0 and w+6=0) instead of multiplying them together?
• This is the zero product property.

It states that if a and b are two real numbers such that ab=0 then a=0 or b=0 (note that both could equal zero in this mathematical usage of the word "or").

The proof is fairly straightforward. Suppose a and b are two numbers whose product is zero. a can either be equal to zero or not equal to zero. If a is equal to zero then we have shown that either a or b is zero (in this case a, it doesn't matter what b is). But if a is not equal to zero, we can take the equation ab=0 and divide both sides by a to get b=0/a=0.

Another way to say this is that if the product of any two numbers is zero, at least one of those numbers MUST be zero. Another way to think of this is that the product of any two non-zero numbers is NEVER zero.
thus in the video, if (w+4) [a number] and (w+6) [another number] multiply to equal zero, then one of them MUST be zero for the equation to be true. thus he writes that either W+4 = 0 or, if that isn't the case, then w+6 must be 0 for the equation to be true.
• At why did Sal write w2+12w+36?
• Sal did the multiplication for ( 6 + w ) ^ 2.
He put the result into standard form for a polynomial which meant that the terms were arranged from highest exponent to lowest.
• So here is my example of a square root equation
so the square root of (9-6) cubed - (4-2)
1st- u solve 9-6 which is 3 then you solve 4-2 which is 2 then u subtract 3 cubed (27) by 2 which is 25 then you solve the square root of 25 which is 5. So did i do it right?
• At when he is factoring, what if there was an equation that was not factorable? Would the answer be no solution?
• Not necessarily. There are several methods that can be used to solve quadratic equations besides using factoring. You will learn about them in later lessons.
(1 vote)
• If two radicals are equal, then the radicands are also equal:
√𝑎 = √𝑏 ⇒ 𝑎 = 𝑏

Just remember that 𝑎, 𝑏 ≥ 0
• √(2x+9) - √(x+1)=√(x+4)
What are the steps to solving for x?
Do you square both sides or what?
• That's a good start. Squaring both sides will leave you with just one radical. Then you can isolate the radical on one side and square again to get rid of it.
• Can anyone help me solve this problem with an explanation? It has two solutions.
(7-(2/x))=√(7/x)