# find x

#### Albert

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

find :$x$

#### anemone

##### MHB POTW Director
Staff member
$(0<x<3), \sqrt {1+x^2}+\sqrt {9+(3-x)^2} =5$

find :$x$
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

Staff member
Here is my solution:

Arrange the equation as follows:

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

Both sides are positive, so squaring yields:

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

Expand squared binomial and collect like terms:

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

Square again:

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

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

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

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

#### Albert

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

find :$x$

the solution using geometry :