Chapter 1.5, Problem 21PSB

a.

To determine

To Calculate: The value of $x$ .

Answer to Problem 21PSB

  $x=2$

Explanation of Solution

Given:

  

  $\begin{array}{l}l\left(\overline{PQ}\right)=x^2+3,l\left(\overline{QR}\right)=4+2x\\ l\left(\overline{PR}\right)=15\text{units}\end{array}$

Concept Used:

A line segment is the line with two end-points

Calculation:

Here, we have

  $\begin{array}{l}l\left(\overline{PQ}\right)=x^2+3,l\left(\overline{QR}\right)=4+2x\\ l\left(\overline{PR}\right)=15\text{units}\\ \because l\left(\overline{PQ}\right)+l\left(\overline{QR}\right)=l\left(\overline{PR}\right)\\ \Rightarrow x^2+3+4+2x=15\\ \Rightarrow x^2+2x-8=0\\ \Rightarrow x^2+4x-2x-8=0\\ \Rightarrow \left(x+4\right)\left(x-2\right)=0\\ \Rightarrow x=-4,x=2\\ \because l\left(\overline{QR}\right)=4+2x=4+2\left(-4\right)=-4\\ \text{But length cannot be in negative}\\ \Rightarrow x\ne -4\\ \therefore x=2\text{is our required value}\text{.}\end{array}$

Conclusion:

  $x=2$

b.

To determine

To Check: Is Q is mid-point of $\overline{PR}$ .

Answer to Problem 21PSB

  $\text{Q is not the mid-point of }\overline{PR}$

Explanation of Solution

Given:

  

  $\begin{array}{l}l\left(\overline{PQ}\right)=x^2+3,l\left(\overline{QR}\right)=4+2x\\ l\left(\overline{PR}\right)=15\text{units}\end{array}$

Concept Used:

A line segment is the line with two end-points

Calculation:

Here, we have

  $\begin{array}{l}l\left(\overline{PQ}\right)=x^2+3,l\left(\overline{QR}\right)=4+2x\\ l\left(\overline{PR}\right)=15\text{units}\\ \because l\left(\overline{PQ}\right)+l\left(\overline{QR}\right)=l\left(\overline{PR}\right)\\ \Rightarrow x^2+3+4+2x=15\\ \Rightarrow x^2+2x-8=0\\ \Rightarrow x^2+4x-2x-8=0\\ \Rightarrow \left(x+4\right)\left(x-2\right)=0\\ \Rightarrow x=-4,x=2\\ \because l\left(\overline{QR}\right)=4+2x=4+2\left(-4\right)=-4\\ \text{But length cannot be in negative}\\ \Rightarrow x\ne -4\\ \therefore x=2\text{is our required value}\text{.}\\ \therefore l\left(\overline{PQ}\right)=2^2+3=7\\ l\left(\overline{QR}\right)=4+2\left(2\right)=8\\ \because l\left(\overline{PQ}\right)\ne l\left(\overline{QR}\right)\\ \text{Q is not the mid-point of }\overline{PR}\text{.}\end{array}$

Conclusion:

  $\text{Q is not the mid-point of }\overline{PR}$