0

When are we allowed to take square roots in an equation?

For example, $x^2 = (x+2)^2$

Are we allowed to take square roots to solve it?

The Pointer
  • 4,182
dugdugdug
  • 1
  • 5
    You can, if you take care with the signs. $a^2=b^2\iff a=\pm b$. – lulu Jan 31 '21 at 13:25
  • but if we do take roots in this example, we get:

    either -x=x+2

    or x=x+2

    meaning x=-1

    or 0=2!

     – dugdugdug Jan 31 '21 at 13:25
  • Ok, so the case $x=x+2$ does not correspond to a possible value of $x$. That's ok. The other case works. – lulu Jan 31 '21 at 13:29
  • Taking square roots is something to be done with care, as others have noted, to make sure that all possibilities have been identified. Care is also needed when squaring - because this can introduce solutions which were not solutions of the original equation. The basic rule is check the answer in the original equation. – Mark Bennet Jan 31 '21 at 13:31

4 Answers4

2

You can take roots, just be aware of the fact that $\sqrt{a^2} = |a|$. So solving $x^2 = (x+2)^2$ is equivalent to solving $|x| = |x+2|$.

UPDATE

Next step is to remember that

  • if $x\le-2$ then the equation becomes $-x = -(x+2)$, which simplifies to $0=-2$ and so has no solutions;
  • if $-2<x<0$ then we get $-x=x+2$, which implies $x=-1$
  • if $x\ge 0$ then we get $x=x+2$, which again has no solutions.

So there is only one solution to the equation at $x=-1$.

gt6989b
  • 54,422
  • so how would you solve |x|=|x+2|? – dugdugdug Jan 31 '21 at 13:30
  • 1
    @dugdugdug We reduce it to the two possible cases $-x=x+2$ or $x=x+2.$ It's the same thing we'd get by factoring $$(x+2)^2-x^2=0$$ as a difference of squares to get $$[(x+2)+x][(x+2)-x]=0,$$ whence $(x+2)+x=0$ or $(x+2)-x=0,$ or more simply, $2x+2=0$ or $2=0.$ Clearly, then, $2x+2=0,$ which is easy to solve. – Cameron Buie Jan 31 '21 at 13:33
  • so it's the equivalent of taking roots but discarding the "wrong" part of the solution, ie discard 2x+2=0 and go with (x+2)+x=0? – dugdugdug Jan 31 '21 at 13:38
  • @dugdugdug please see the update – gt6989b Jan 31 '21 at 15:58
1

Let's work through equivalences:$$\begin{align}x^2=(x+2)^2&\iff x=x+2\lor -x=x+2\\&\iff-x=x+2\\&\iff x=-1.\end{align}$$

J.G.
  • 115,835
0

If |a|= |b| then either a= b or a= -b. So if |x|= |x+ 2| either x= x+ 2 or x= -(x+ 2)= -x- 2. If x= x+ 2 then, subtracting x from both sides, 0- 2 which, of course, is not true. x is not equal to x+ 2. If x= -x- 2, adding x to both sides, 2x= -2 so x= -1. You can (and should) check back in the original problem: if x= -1, x^2= (-1)^2= 1 while (x+ 2)^2= (-1+ 2)^2= 1^2= 1. Yes, they are equal!

HallsofIvy
  • 86
  • 1
    The reason for posing this question is as a student, you follow what teachers say, ie whatever you do to one side of an equation, you do to the other.

    But by taking roots, you end up with 0=2, which clearly doesn't make sense.

    So if the student asks: "when can you take roots?", would the correct answer be, "always but be careful"?

     – dugdugdug Jan 31 '21 at 13:40
0

This can be solved instead as $$(x+2)^2-x^2=0=(x+2+x)(x+2-x)=4(x+1)$$

Using the difference of two squares in this way saves the taking of square roots, and can be less error-prone as well as making it clear what is going on.

The "missing solution" from the quadratic is identified by setting $x=\frac 1y$ so that $$\frac {1}{y^2}=\left(\frac 1y+2\right)^2$$ or $$1=\left(1+2y\right)^2$$ which has the solution $y=0$ and indicates a "missing solution at infinity". $y=-1$ corresponds to the original solution for $x$.

Mark Bennet
  • 100,194
  • so would the method for solving x^2=(x+2)^2 be the same as solving:

    x^100=(x+2)^100

     – dugdugdug Jan 31 '21 at 13:50
  • @dugdugdug - not really, or perhaps yes. To use the same method you need to know that $y^{100}-x^{100}$ is the product of $100$ linear factors of the form $(y-\alpha x)$ where $\alpha$ is a $100^{th}$ root of unity. Each root of unity gives a linear factor, and each linear factor potentially gives a solution. In the concrete case you give, still have the missing solution at infinity (where the factorisation contains a constant factor). The solution can always be done by taking $100^{th}$ roots directly, provided all the roots are considered. – Mark Bennet Jan 31 '21 at 14:00