Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
 A: Let $y=\sqrt{x}$, therefore $y^2=x$
$y^2-8y+7=0$ therefore $(y-7)(y-1)=0$ hence $y=1,7$.
Therefore $x=1,49$. These are indeed the only solutions.
A: Isolate the square root, then square both sides:
\begin{align*}
     x-8\sqrt x+7&=0\\
     \implies x+7&=8\sqrt x\\
     \implies (x+7)^2 &= 64 x\\
     \implies x^2 -50x + 49&=0\\
     \implies (x-49)(x-1) &=0\\
     \implies x = 49 \text{ or } x&=1,
\end{align*}
which are the two solutions.
A: If
$x - 8\sqrt x + 7 = 0, \tag 1$
then
$x + 7 = 8\sqrt x; \tag 2$
then
$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$
thus
$x^2 - 50x + 49 = 0, \tag 4$
which factors as
$(x - 1)(x - 49) = 
x^2 - 50x + 49 = 0; \tag 5$
thus
$x = 1 \; \text{or} \; x = 49; \tag 6$
it is now a simple matter to check that $1$, $49$ obey (1).
A: Let $\sqrt{x}=a$
$$a^2-8a+7=0$$
Find the factor
$$(a-7)(a-1)=0$$
Then you have
$$a=7\text{ and }a=1$$
Remember that $a=\sqrt{x}$, then
$$x=1 \text{ or } x=49$$
A: Why did things get compilicated?
If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get
$y^2 - 8y +7 =0$ and that should not be complicated.
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Factor or use the quadratic formula so 
$(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$.
Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\pm 6}2=4\pm 3=1,7$; so $y=7$ or $y =1$.
So $x = y^2$ so $x = 7^2 = 49$ or $x = 1$.  Thus are the two solutions.
Check: 
If $x = 49$ then $x - 8\sqrt x +7 = 49 - 8*7 + 7 = 49- 56 + 7 = 0$.
And if $x = 1$ then $x - 8\sqrt x + 7 = 1-8*1 + 7 = 0$.
A: $\displaystyle x-8\sqrt{x}+7=0$
or, $\displaystyle (\sqrt{x} -1)(\sqrt{x} -7)=0$
Hence, $x = 1$ and $49$
A: hint
It can be written as
$$8x-7x-8\sqrt{x}+7=$$
$$8\sqrt{x}(\sqrt{x}-1)-7(\sqrt{x}-1)(\sqrt{x}+1)=$$
$$(\sqrt{x}-1)\Bigl(8\sqrt{x}-7(\sqrt{x}+1)\Bigr)=$$
$$(\sqrt{x}-1)(\sqrt{x}-7)=0$$
thus...
A: It turns out you made the right substitution $ x = y^{2} $. Using that substitution, $$ y^2 - 8y + 7 $$
Observe that the above function is factorable:
$$ y^2 - 8y + 7 = (y - 7)(y - 1) = 0$$
The solutions for $y$ are $y = 7$ and $y = 1$. Back-substituting to $x$, we have that 
$$ x = 49; x = 1 $$
