Finding all real solutions of $x-8\sqrt{x}+7=0$

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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 $x - 8\sqrt x + 7 = 1-8*1 + 7 = 0$.

Answer 5

$\displaystyle x-8\sqrt{x}+7=0$

or, $\displaystyle (\sqrt{x} -1)(\sqrt{x} -7)=0$

Hence, $x = 1$ and $49$

Answer 6

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...

Answer 7

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 $$x=1 \text{ or } x=49$$

Answer 8

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

$$ x = 49; x = 1 $$