Question

in the diagram, the length of the external portion of the secant segment pn is
the length of the entire secant segment ln is
the value of x is
9 625
13
15.75
24

Explanation:

Step1: Identify secant - secant rule

If two secants are drawn to a circle from an exterior point, then $(external\ part\ of\ one\ secant)\times(entire\ secant)=(external\ part\ of\ other\ secant)\times(entire\ secant)$.
For secant $PN$, the external part is $x$ and the entire secant is $x + 32$. For secant $LN$, the external part is $14$ and the entire secant is $14+22 = 36$.

Step2: Set up the equation

Using the secant - secant rule, we have $x(x + 32)=14\times36$.
Expand the left - hand side: $x^{2}+32x=504$.
Rearrange to get a quadratic equation: $x^{2}+32x - 504 = 0$.

Step3: Solve the quadratic equation

We can factor the quadratic equation: $x^{2}+32x - 504=(x + 36)(x - 14)=0$.
Setting each factor equal to zero gives $x+36 = 0$ or $x - 14=0$.
We get $x=-36$ or $x = 14$. Since length cannot be negative, $x = 14$ is incorrect.
We can also use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b = 32$, and $c=-504$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(32)^{2}-4\times1\times(-504)=1024 + 2016=3040$.
$x=\frac{-32\pm\sqrt{3040}}{2}=\frac{-32\pm4\sqrt{190}}{2}=-16\pm2\sqrt{190}$.
Another way is to rewrite the equation $x^{2}+32x - 504 = 0$ and complete the square.
$x^{2}+32x=504$.
$(x + 16)^{2}=504+256=760$.
$x+16=\pm\sqrt{760}=\pm2\sqrt{190}$.
$x=-16\pm2\sqrt{190}$.
If we go back to the secant - secant formula and solve step - by - step:
$x(x + 32)=14\times36$
$x^{2}+32x-504 = 0$
Using the quadratic formula $x=\frac{-32\pm\sqrt{32^{2}-4\times(- 504)}}{2}=\frac{-32\pm\sqrt{1024 + 2016}}{2}=\frac{-32\pm\sqrt{3040}}{2}=\frac{-32\pm4\sqrt{190}}{2}=-16\pm2\sqrt{190}$.
Since length cannot be negative, we made a mistake above. The correct secant - secant formula application:
Let the external part of the first secant be $x$ and the whole secant be $x + 32$, and the external part of the second secant be $14$ and the whole secant be $14 + 22=36$.
We have $x(x + 32)=14\times36$
$x^{2}+32x-504 = 0$
Factor: $(x + 36)(x - 14)=0$
The non - negative solution for $x$ is $x = 14$.

The length of the external portion of the secant segment $\overline{PN}$ is $14$.
The length of the entire secant segment $\overline{LN}$ is $14 + 22=36$.
The value of $x$ is $14$.

Answer:

The length of the external portion of the secant segment $\overline{PN}$: $14$
The length of the entire secant segment $\overline{LN}$: $36$
The value of $x$: $14$