Why can you not square each individual term in an equation like $\sqrt{11} +\sqrt{44}= \sqrt{99}$?

Question

Given the equation

$\sqrt{11} +\sqrt{44}= \sqrt{99}$

Why is squaring each individual term not allowed? Doing so we get 11+44=99 which is incorrect. Is it because $\sqrt{11} +\sqrt{44}$ is considered a term grouped by addition and therefore to be treated as a single entity instead of two parts?

Answer 1

It is because 'squaring' is not linear.

That means that the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ has not the property that $f(a+b)=f(a)+f(b)$ since $f(a+b)=(a+b)^2=a^2+2ab+b^2\neq a^2+b^2=f(a)+f(b)$.

You can square both sides, but you have to be more specific.

If you square (what can be done since both sides are positive) we get:

$(\sqrt{11}+\sqrt{44})^2=(\sqrt{99})^2$

Which leads to:

$11+44+2\sqrt{11}\sqrt{44}=99$

So $2\sqrt{11\cdot 44}=44$

$\sqrt{484}=22\checkmark$

Answer 2

In an equation, you are allowed to do the same operation to both sides. So squaring the whole left side gives you $(\sqrt{11}+\sqrt{44})^2$. And "unfortunately", this is not the same as $\sqrt{11}^2+\sqrt{44}^2$. So yes, in a sense, you have to think of the entire left-hand-side as a single term in this respect.

Answer 3

Because $(a+b)^2=a^2+2ab+b^2$. There’s a cross term $2ab$.

Answer 4

You can think it Geometrically. Assume you cut three pieces of paper in square shape of 2 cm , 3 cm and 5 cm. According to you Both small paper should fully cover bigger one. but if you Observe , there is some area left which can easily be seen is $2(2.3)$[two times a rectangle]

Answer 5

The simplest reason that it isn't allowed is that it simply doesn't work. If you take the numbers $\sqrt{11}$ and $\sqrt{44}$, then add them, and then square the result, you do not get $11+99$, you get $11+44+2\sqrt{11\cdot 44}=11+44+2\cdot 22=99$.

We can get the correct result as shown above by using a different method. It is the distributive property, where we take each piece and multiply it out. To do the above calculation more carefully:

$$(\sqrt{11}+\sqrt{44})^2=(\sqrt{11}+\sqrt{44})\cdot (\sqrt{11}+\sqrt{44})$$$$= (\sqrt{11}+\sqrt{44})\cdot (\sqrt{11})+(\sqrt{11}+\sqrt{44})\cdot (\sqrt{44})$$ $$=11+44+2\sqrt{11\cdot 44}=11+44+(2\cdot 22)=99$$

This works for any numbers at all, not just 11 and 44. For instance:

$$(\sqrt{4}+\sqrt{49})^2=(\sqrt{4}+\sqrt{49})\cdot (\sqrt{4}+\sqrt{49})$$$$= (\sqrt{4}+\sqrt{49})\cdot (\sqrt{4})+(\sqrt{4}+\sqrt{49})\cdot (\sqrt{49})$$ $$=4+49+2\sqrt{4\cdot 49}=4+49+(2\cdot 14)=81$$

Which you may note agrees with the more obvious: $(2+7)^2=81$. If we assumed that the square can "distribute" as in your conjecture, we would get the incorrect statement $(2+7)^2=4+49=53$.

So even though it looks as though it may be reasonable for the square to distribute, it cannot do that, or else we would have nonsense.

This should, I hope, make it a bit more clear that a) the distributive property (for multiplying) is very useful and cool, and a very interesting property for multiplication to have AND b) the rules for squaring and other powers are actually direct consequences of the distributive property. Squaring means multiplying a number by itself, and the distributive property determines what that must be.