Chapter 7, Problem 14CR

To determine

To solve: The system of equations graphically.

Answer to Problem 14CR

The solution of the system is $(-\sqrt{2},5+4\sqrt{2})\mathrm{and}(\sqrt{2},5-4\sqrt{2})$ .

Explanation of Solution

Given information: The given equations are:

  $\begin{array}{l}y=2x^2-4x+1(i)\\ y=x^2-4x+3(ii)\end{array}$ .

Calculation: The given equations are:

  $\begin{array}{l}y=2x^2-4x+1(i)\\ y=x^2-4x+3(ii)\end{array}$ .

To solve the system by graphically:

From equation $(i)$

  $\begin{array}{l}y=2x^2-4x+1\\ \mathrm{put}x=0\\ $ y=2{(0)}^2-4(0)+1$y=0-0+1=1\\ \mathrm{put}x=1\\ $ y=2{(1)}^2-4(1)+1$y=2-4+1=3-4=-1\end{array}$

From equation $(ii)$

  $\begin{array}{l}y=x^2-4x+3\\ \mathrm{put}x=0\\ $ y={(0)}^2-4(0)+3$y=0-0+3=3\\ \mathrm{put}x=1\\ $ y={(1)}^2-4(1)+3$y=1-4+3=4-4=0\end{array}$

Plot these coordinate in graph and find the solution:

  

The solution from the graph is $(-\sqrt{2},5+4\sqrt{2})\mathrm{and}(\sqrt{2},5-4\sqrt{2})$ .

Check:

Put $(-\sqrt{2},5+4\sqrt{2})\mathrm{and}(\sqrt{2},5-4\sqrt{2})$ equation $(i)\mathrm{and}(ii)$ :

  $\begin{array}{l}y=2x^2-4x+1\\ $ 5+4\sqrt{2}=2{(-\sqrt{2})}^2-4(-\sqrt{2})+1$5+4\sqrt{2}=2(2)+4\sqrt{2}+1\\ 5+4\sqrt{2}=4+4\sqrt{2}+1\\ 5+4\sqrt{2}=5+4\sqrt{2}\\ \mathrm{Satisfied}.\\ y=x^2-4x+3\\ $ 5+4\sqrt{2}=2{(-\sqrt{2})}^2-4(-\sqrt{2})+3$5+4\sqrt{2}=2+4\sqrt{2}+3\\ 5+4\sqrt{2}=5+4\sqrt{2}\\ \mathrm{Satisfied}.\\ y=2x^2-4x+1\\ $ 5-4\sqrt{2}=2{(\sqrt{2})}^2-4(\sqrt{2})+1$5-4\sqrt{2}=2(2)-4\sqrt{2}+1\\ 5-4\sqrt{2}=4-4\sqrt{2}+1\\ 5-4\sqrt{2}=5-4\sqrt{2}\\ \mathrm{Satisfied}.\\ y=x^2-4x+3\\ $ 5-4\sqrt{2}=2{(\sqrt{2})}^2-4(\sqrt{2})+3$5-4\sqrt{2}=2-4\sqrt{2}+3\\ 5-4\sqrt{2}=5-4\sqrt{2}\\ \mathrm{Satisfied}.\end{array}$

Thus the solution of the system is $(-\sqrt{2},5+4\sqrt{2})\mathrm{and}(\sqrt{2},5-4\sqrt{2})$ .