Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$

Question

I solved the following equation, and I just want to be sure I did it right.

This is the procedure:

$$ \sqrt{x-4}-\sqrt{x-5}+1=0\\ \sqrt{x-4}=\sqrt{x-5}-1\\ \text{squaring both sides gives me:}\\ (\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\ x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\ x-4=x-5-\sqrt{x-5}+1\\ x-4=x-4-\sqrt{x-5}\\ \text{substracting x, and adding 4 to both sides}\\ 0=-\sqrt(x-5)\\ \text{switching both sides}\\ \sqrt{x-5}=0\\ \text{sqaring both sides}\\ x-5=0\\ x=5\\ \text{When I place 5 in the equation, I get:}\\ \sqrt{5-4}-\sqrt{5-5}+1=0\\ \sqrt{1}-\sqrt{0}+1=0\\ 1-0+1=0\\ 2=0\\ \text{this means that the equation dosent have any solution, right??}\\ $$ Any advice and suggestion is helpful.

Thanks!!!

$$(\sqrt{x}-1)^2 = x - 2\sqrt{x} + 1$$ and you seem to have dropped the 2... — gt6989b, Jan 02 '14 at 15:40

score 5 · Accepted Answer · answered Jan 02 '14 at 15:44

5

Easiest way to see it, is take both square roors to the other side and square, to get $$ \begin{split} 1 &= (x-5)-(x-4) - 2\sqrt{(x-5)(x-4)}\\ 2 + 2\sqrt{(x-5)(x-4)} &= 0\\ 1 + \sqrt{(x-5)(x-4)} &= 0\\ \end{split} $$ but $\sqrt{\ldots} \geq 0$ so this is impossible...

answered Jan 02 '14 at 15:44

gt6989b

54,422

Can you explain me, how do you know that $\sqrt{...} < 0$. Thanks. – depecheSoul Jan 02 '14 at 15:47
2

@depecheSoul : He doesn't, his last equation implies $\sqrt{...} = -1$ which is impossible. – user88595 Jan 02 '14 at 15:56
1

$\sqrt4=\sqrt(\pm 2)^2=\pm 2$ – Real Hilbert Jan 02 '14 at 16:00
1

@RealHilbert If you plug in any real value for $x$ into $\sqrt{(x-5)(x-4)}$, you will never get anything negative out... – gt6989b Jan 02 '14 at 16:02
Got it thanks everybody!! – depecheSoul Jan 02 '14 at 16:02
1

check $x=\frac{9+\sqrt5}{2}$ you may get -1 – Real Hilbert Jan 02 '14 at 16:12
1

@RealHilbert when a equation involving the function $\sqrt{x}$ is given, it's often implicit that one of the positive or negative value is chosen for all square roots. In both cases, it's correct to conclude that the square root appearing in last equation is positive. – pppqqq Jan 02 '14 at 16:14
@RealHilbert We are not dealing with complex curves, there are no branches. $\sqrt{x} > 0$ by definition... – gt6989b Jan 02 '14 at 19:09

score 4 · Answer 2 · answered Jan 02 '14 at 15:40

4

Note that the expansion of $(\sqrt{x - 5} - 1)^2 = (\sqrt{x - 5})^2 - {\bf 2}\sqrt {x - 5} + 1$.

But that correction doesn't change the fact that indeed, no solution exists.

answered Jan 02 '14 at 15:40

amWhy

209,954

Needs another UV+1 – Jan 02 '14 at 18:36
@amWhy: Agreed, needs another UV +1 – Amzoti Jan 03 '14 at 14:26

score 3 · Answer 3 · answered Jan 02 '14 at 15:46

It is easily seen that the equation $\sqrt{x-4}-\sqrt{x-5}=-1$ has no solutions since $\sqrt{x-4}>\sqrt{x-5}$ for all $x>=5$.

Your method is correct but you made a mistake in going from step $3$ to step $4$. After correcting that, you will still reach the same conclusion that there are no solutions.

score 2 · Answer 4 · answered Jan 02 '14 at 16:14

2

The square root function is increasing on $[0,\infty)$, so $$\sqrt{x-4} - \sqrt{x-5} \ge 0 $$ for all $x\ge 5.$ Hence the expression can never equal -1.

answered Jan 02 '14 at 16:14

ncmathsadist

49,383

score 1 · Answer 5 · answered Jan 02 '14 at 15:49

1

Hint $\ \ \sqrt{z+1}+\sqrt{z}\, =:\, y\ $ times $\,(\sqrt{z+1}-\sqrt{z}\,=\,-1) \ \Rightarrow\ 1 = -y\,$ contra $\, y > 0\ \ \ $ QED

answered Jan 02 '14 at 15:49

Bill Dubuque

272,048

score 1 · Answer 6 · answered Jan 02 '14 at 16:09

1

As $(x-4)-(x-5)=1$

$\displaystyle(\sqrt{x-4}-\sqrt{x-5})(\sqrt{x-4}+\sqrt{x-5})=1$

As $\displaystyle\sqrt{x-4}-\sqrt{x-5}=-1, \sqrt{x-4}+\sqrt{x-5}=-1$

Adding we get $\displaystyle\sqrt{x-4}=-1$ which is impossible as $\sqrt{x-1}\ge0$ for real $x$

answered Jan 02 '14 at 16:09

lab bhattacharjee

274,582

score 1 · Answer 7 · answered Jan 02 '14 at 17:00

Set $a=\sqrt{x-4}$ and $b=\sqrt{x-5}$ for $-2\leq x\leq +2$. Equivalently we have $a^2=x-4$, $b^2={x-5}$ for $-2\leq x\leq +2$. It's implies for $-2\leq x\leq +2$, \begin{align} a-b=-1 \implies & (a-b)(a+b)=-1\cdot(a+b)\\ \implies & a^2-b^2=-(a+b)\\ \implies & (x-4)-(x-5)=-(a+b)\\ \implies & -1=-(a+b)\\ \implies & a+b=+1.\\ \end{align} The same calculations and similar implications $a+b=+1\implies a-b=+1$. Then $ x $ is a root of $a+b=-1$, if only if, \begin{cases} a+b=&+1\\ a-b=&-1\\ \end{cases} Here, $a=0$ and $b=1$ implies $a^2=x-4=0$ and $b^2={x-5}=1$. In both cases, $x\in\!\!\!\!\!/[-2,+2]$. So the equation $$ \sqrt{x-4}-\sqrt{x-5}+1=0 $$ has no solution.

Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$

7 Answers7