Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$

Question

Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?

Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$

Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$

Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$

We have removed one square root.

Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?

Simplify: $2x-5 = 2\sqrt{(x-1)} + x$

Simplify more: $x-5 = 2\sqrt{(x-1)}$

Now do the "square root" thing again:

Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$

Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$

Square root removed

Thank you in advance for your help

Answer 1

I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.

Answer 2

$$2x-5 = (1 + \sqrt{x-1})^2$$

to expand RHS use this formula or simple mulipty it with itself(to do square). formula is:

$(a+b)^2=a^2+b^2+2\times a\times b$

so your expansion will be $$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$ $$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$ $$2x-5 = x+2\sqrt{x-1}$$ $$x-5 = 2\sqrt{x-1}$$ now you have your way.

Answer 3

You can solve the square of a sum by writing out the square as a product of two sums and writing out the multiplication for each pair of terms.

$$(1+\sqrt{x-1})^2 =\\ (1+\sqrt{x-1})(1+\sqrt{x-1})=\\ 1\times1 + 1\times\sqrt{x-1}+\sqrt{x-1}\times1+\sqrt{x-1}\times\sqrt{x-1}=\\ 1+\sqrt{x-1}+\sqrt{x-1}+x-1=\\ 2\sqrt{x-1}+x$$

Answer 4

It is an identity: $$(1+x)^2=1+2x+x^2$$ How? Well, consider this: $$\begin{equation*} \begin{split} (1+x)^2&=(\color{blue}{1}+\color{red}{x})(1+x)\\ &=\color{blue}{1}(1+x)+\color{red}{x}(1+x)\\ &=1+x+x+x^2\\ &=1+2x+x^2\\ \end{split} \end{equation*}$$ In general, $(a+b)^2=a^2+2ab+b^2$, which you can try to deduce yourself.

Hope this helps. Ask anything if not clear :)

Answer 5

$\sqrt{2x-5} - \sqrt{x-1} = 1$

Let $\sqrt{2x-5} + \sqrt{x-1} = y$

Multiplying, we get

$(2x-5) - (x-1) = y$

$y = x - 4$

\begin{align} \sqrt{2x-5} + \sqrt{x-1} &= x - 4 \\ \sqrt{2x-5} - \sqrt{x-1} &= 1 & \text{subtract}\\ \hline 2\sqrt{x-1} &= x-5 \\ 4x-4 &= x^2 - 10x + 25 \\ x^2 -14x + 29 &= 0 \\ x &= 7 \pm 2 \sqrt 5 \end{align}

Answer 6

To get rid of the square root, denote: $\sqrt{x-1}=t\Rightarrow x=t^2+1$. Then: $$\sqrt{2x-5} - \sqrt{x-1} = 1 \Rightarrow \\ \sqrt{2t^2-3}=t+1\Rightarrow \\ 2t^2-3=t^2+2t+1\Rightarrow \\ t^2-2t-4=0 \Rightarrow \\ t_1=1-\sqrt{5} \text{ (ignored, because $t>0$)},t_2=1+\sqrt{5}.$$ Now we can return to $x$: $$x=t^2+1=(1+\sqrt5)^2+1=7+2\sqrt5.$$