Question

Answer my question which is in attachment

No spamming

Irrelevant answer at a time deleted ​

Answer 1

Topic

• Coordinate Geometry

It is a part of Mathematics that uses coordinates of points.

Solution

① Rewriting the terms.

Given,

$\dfrac{\sqrt{x^{4}+x^{2}+2x+1} +\sqrt{x^{4}-2x^{3}+5x^{2}-4x+1}}{x}$

$=\sqrt{x^{2}+1+\dfrac{2}{x} +\dfrac{1}{x^{2}} } +\sqrt{x^{2}-2x+5-\dfrac{4}{x} +\dfrac{1}{x^{2}} }$

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

② Interpreting as a distance formula.

Let's interpret the above as the distance between two points.

Consider the two square roots.

➊ $\sqrt{x^{2}+(\dfrac{1}{x} +1)^{2}}$

➋ $\sqrt{(x-1)^{2}+(\dfrac{1}{x} -2)^{2}}$

③ What do they require?

In ➊ we find the following.

Distance between a point $(x,\dfrac{1}{x} )$ and a point $(0,-1)$ is required.

In ➋ we find the following.

Distance between a point $(x,\dfrac{1}{x} )$ and a point $(1,2)$ is required.

So, let's call the given points $P(x,\dfrac{1}{x} )$, $A(0,-1)$, $B(1,2)$.

④ Least Distance

The given square roots are equivalent to the following.

$\overline{AP}=\sqrt{x^{2}+(\dfrac{1}{x} +1)^{2}}$

$\overline{BP}=\sqrt{(x-1)^{2}+(\dfrac{1}{x} -2)^{2}}$

Then the given expression is $\overline{AP} +\overline{BP}$, and the length of a line passing through points is the least distance. (Attachment included.)

$\overline{AP} +\overline{BP}$

$\geq \overline{AB}$

$=\sqrt{(1-0)^{2}+(2+1)^{2}}$

$=\sqrt{10}$

⑤ At which value of $x$ is the distance the least?

The minimum value is $\sqrt{10}$, if the three points lie on the same line(or, colinear.)

The slope of $\overline{AB}$ is the following.

$\implies \dfrac{(-1)-2}{0-1}=3$

The slope of $\overline{AP}$ is the following.

$\implies \dfrac{-1-\frac{1}{x} }{0-x} =\dfrac{1+\frac{1}{x} }{x}$

Three points are colinear if any slope of two points is equal.

$\implies \dfrac{1+\frac{1}{x} }{x}=3$

$\implies 1+\dfrac{1}{x} =3x$

$\implies 3x^{2}-x-1=0$

$\implies x=\dfrac{1+\sqrt{13} }{6}$

Conclusion

At $x=\dfrac{1+\sqrt{13} }{6}$, the least value is $\sqrt{10}$.