Problems in manipulating these 4 radical equations

[mark2142]

Homework Statement

I am having trouble in understanding this.

Relevant Equations

1. $\sqrt{2x+1}=x+1$
$(x+1)^2=(2x+1)$

2.$\sqrt{x+1}=8-2x$
$(8-2x)^2= (x+1)$

3. $\sqrt{5-x}=x-3$
$(x-3)^2=(5-x)$

4.$\sqrt{\sqrt {x-5}+x}= 5$
$5^2= \sqrt{x-5}+x$
$25-x=\sqrt{x-5}$
$(25-x)^2=(x-5)$

In order to write next step in all four equations above l used the definition of radicals. $\sqrt a=b$ means $b^2=a$. Squaring both sides also works. I don’t know if it’s right. I mean I read that $(\sqrt a)^2=a$. But I don’t know if we can apply this on expressions.
Main problem is if we look at the equation $x^2=5$ it doesn’t mean the next equation is $x= \sqrt 5$ but $x=+-\sqrt 5$. It seems I can’t use the definition everywhere.
In book it’s done like this:
$x^2-5=0$.
$x^2- (\sqrt 5)^2=0$
$(x- \sqrt 5)(x+ \sqrt 5)=0$
$x =+- \sqrt 5$.
And I get this but definition of radical should also give me same answer.

Other thing is $\sqrt {(-4)^2}= 4 \neq -4$ but when we do square of square roots in equations we simply write, say in 1. $(\sqrt {2x+1})^2= 2x+1$. I mean can we cancel the powers or not?
Now look at this equation : $2x-1=- \sqrt {2-x}$.
We can square but we can’t use the definition of radicals without shifting -ve to left hand side.
This whole business is somewhat confusing. It feels like I don’t understand radical equations fully or do I?
Also if you give any explanation please use simple words. I am not very good at maths.

 

[pasmith]

$\sqrt{}$ means the positive root; $\sqrt{x^2} = |x|$.

If $x^2 = a > 0$ then there are two possibilities: either $x = \sqrt{a}$ or $x = -\sqrt{a}$. Squaring both sides of either of these takes you back to $x^2 = a$.

EDIT: If we confine attention to real numbers, then $(\sqrt{x})^2 = x$ because $x$ must be positive in order for us to take its root.

 

[mark2142]

means the positive root; x2=|x|.

Yes, So we can’t just cancel the powers. $\sqrt {(-4)^2}= 4$.
How come then $(\sqrt{2x+1})^2= \sqrt{(2x+1)^2}=2x+1$?

 

[pasmith]

Yes, So we can’t just cancel the powers. $\sqrt {(-4)^2}= 4$.
How come then $(\sqrt{2x+1})^2= \sqrt{(2x+1)^2}=2x+1$?

If we are dealing with real numbers, we must have $2x + 1 \geq 0$ in order to take the root; thus $|2x + 1| = 2x + 1$.

 

[mark2142]

If we are dealing with real numbers, we must have $2x + 1 \geq 0$ in order to take the root; thus $|2x + 1| = 2x + 1$.

That I missed. Good point. And now I understand that $(\sqrt{2x+1})^2= ({2x+1})^{\frac 12(2)}= 2x+1$.

If x2=a>0 then there are two possibilities: either x=a or x=−a. Squaring both sides of either of these takes you back to x2=a.

In this part ( $x^2=5$) why can’t I just use the definition of radicals ?

 

[MidgetDwarf]

The main idea when solving equations when we are dealing with real numbers, is that we first want to analyze when, if any extraneous solutions occur.

For example, we cannot divide by zero, log function is defined for numbers larger than o, we cannot take the square root of negative numbers, just to name a few scenarios.

Now, the definition of a square root of a real number, requires that the number be greater than or equal to 0.

So the first step one must do, is use the definition the square root of a positive real number, and consider
2x + 1 ≥ 0. Solving this inequality, we want the x we solve for in our last step to be x ≥ -1/2.

Then proceed using the algebra you know, solve for x.

 

[pasmith]

In this part ( $x^2=5$) why can’t I just use the definition of radicals ?

The rule $$(x^a)^b = x^{ab} = (x^b)^a$$ always holds for integer $a$ and $b$, but otherwise only holds for nonnegative real $x$. Dropping this restriction creates issues: $$((-1)^2)^{1/2} = 1 \neq -1 = ((-1)^{1/2})^2$$ Thus, if you are restricting $x$ to be real and nonnegative, then there is no problem with $$x^2 = 5\quad\Rightarrow\quad (x^2)^{1/2} = 5^{1/2}\quad\Rightarrow\quad x = \sqrt{5}.$$ If you don't have that restriction and also allow $x < 0$ then the above does not hold: it could instead be the case that $x = -\sqrt{5}$.

 

[mark2142]

The rule $$(x^a)^b = x^{ab} = (x^b)^a$$ always holds for integer $a$ and $b$, but otherwise only holds for nonnegative real $x$. Dropping this restriction creates issues: $$((-1)^2)^{1/2} = 1 \neq -1 = ((-1)^{1/2})^2$$ Thus, if you are restricting $x$ to be real and nonnegative, then there is no problem with $$x^2 = 5\quad\Rightarrow\quad (x^2)^{1/2} = 5^{1/2}\quad\Rightarrow\quad x = \sqrt{5}.$$ If you don't have that restriction and also allow $x < 0$ then the above does not hold: it could instead be the case that $x = -\sqrt{5}$.

I am given an equation to solve and I am unable to understand why definition of root doesn’t work both ways. Namely if $\sqrt 5 = x$ means $x^2=5$ then $x^2=5$ doesn’t means $\sqrt 5=x$ It has to do with root and square. Square having unique value but root gives two values. But I don’t understand it clearly. Can you explain in those terms?

 

[pasmith]

Given $x \in \mathbb{R}$, $x^2$ is uniquely determined: the laws of arithmetic would not hold if there were two or more possible values for $x^2$. But given $c > 0$ there are always two numbers such that $x^2 = c$, because the laws of arithmetic require $(-x)^2 = x^2$.

By convention, for $c \geq 0$ we define $\sqrt{c}$ to be the nonnegative root. If $x^2 = c$ then either $x = \sqrt{c} \geq 0$ or $x = -\sqrt{c} \leq 0$.