MHB What is the value of x for the inequality problem?

[Albert1]

$(0<x<3), \sqrt {1+x^2}+\sqrt {9+(3-x)^2} =5$

find :$x$

 

[anemone]

$(0<x<3), \sqrt {1+x^2}+\sqrt {9+(3-x)^2} =5$

find :$x$

Spoiler

If we let $x=\tan y$ and notice that $1+(\tan y)^2=\sec^2 y$, we get:

$\sqrt {1+(\tan y)^2}+\sqrt {9+(3-\tan y)^2} =5$

$\sec y+\sqrt {9+(3-\tan y)^2} =5$

$\sqrt {9+(3-\tan y)^2} =5-\sec y$

Square both sides of the equation and simplify, we have:

$-3\tan y =4-5\sec y$

Squaring again and use the identity $1+(\tan y)^2=\sec^2 y$ we obtain:

$(4\sec y-5)^2=0$

In other words, $\sec y =\dfrac{5}{4}$ and this implies $\tan y= \dfrac{3}{4}$ or $x=\dfrac{3}{4}$.

 

[MarkFL]

Here is my solution:

Spoiler

Arrange the equation as follows:

$$\sqrt{9+(3-x)^2}=5-\sqrt{1+x^2}$$

Both sides are positive, so squaring yields:

$$9+(3-x)^2=25-10\sqrt{1+x^2}+1+x^2$$

Expand squared binomial and collect like terms:

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

Square again:

$$16+24x+9x^2=25\left(1+x^2 \right)$$

$$16x^2-24x+9=0$$

$$(4x-3)^2=0$$

$$x=\frac{3}{4}$$

 

[Albert1]

$(0<x<3), \sqrt {1+x^2}+\sqrt {9+(3-x)^2} =5$

find :$x$

Spoiler

the solution using geometry :

View attachment 1847