Chapter 9.6, Problem 35E

To determine

To calculate the solution of the given set of equations.

Answer to Problem 35E

The solution is $x\approx -2.53,0.19$ and $y\approx 1.04,1.6$

Explanation of Solution

Given information:

  $\begin{array}{l}y=2{(}^4-1\\ y=3x^2+8x\end{array}$

Formula used:

Approximation

Calculation:

Equation 1 is $y=2{(4)}^x-1$

Equation 2 is $y=3x^2+8x$

Graphing the two equations,

  

Concluding from the graph, the set of equation has two solution, which lines between $x=0$ and $x=1$ ,

  $x=-2$ and $x=-3$ ,

Substituting equation 2 in equation 1,

  $2{(4)}^x-1=3x^2+8x$

Shifting the values,

  $3x^2+8x-2{(4)}^x+1=0$

Since, this equations cannot be solved algebraically, let,

  $f(x)=3x^2+8x-2{(4)}^x+1$

Evaluating f(x) between $x=0$ and $x=1$ ,

For f(0.1), substituting the value,

  $f(0.1)=3{(0.1)}^2+8(0.1)-2{(4)}^{0.1}+1$

On solving,

  $f(0.1)=-0.47$

For f(0.2), substituting the value,

  $f(0.2)=3{(0.2)}^2+8(0.2)-2{(4)}^{0.2}+1$

On solving,

  $f(0.2)=0.081$

Since, f(0.1)<0 and f(0.2)<0, the zero is between 0.1 and 0.2

Since f(0.1) is closer to zero than f(0.1), so decrease the guess and evaluate f(0.2)

For f(0.19) substituting the values,

  $f(0.19)=3{(0.19)}^2+8(0.19)-2{(4)}^{0.19}+1$

On solving,

  $f(0.19)=0.026$

For f(0.18) substituting the values,

  $f(0.18)=3{(0.18)}^2+8(0.18)-2{(4)}^{0.18}+1$

On solving,

  $f(0.18)=0.030$

Because, f(0.19) is the closest to 0

Therefore, $x\approx 0.19$

For y, substituting the values,

  $y=3{(0.19)}^2+8(0.19)$

On solving,

  $y\approx 1.6$

Evaluating f(x) between $x=-2$ and $x=-3$ ,

For f(-2.5), substituting the value,

  $f(-2.5)=3{(-2.5)}^2+8(-2.5)-2{(4)}^{-2.5}+1$

On solving,

  $f(-2.5)=-0.31$

For f(-2.4), substituting the value,

  $f(-2.4)=3{(-2.4)}^2+8(-2.4)-2{(4)}^{-2.4}+1$

On solving,

  $f(-2.4)=-0.99$

Since, f(-2.4)<0 and f(-2.5)<0, the zero is between -2.4 and -2.5

Since f(-2.5) is closer to zero than f(-2.4), so decrease the guess and evaluate f(2.5)

For f(-2.51) substituting the values,

  $f(-2.51)=3{(-2.51)}^2+8(-2.51)-2{(4)}^{-2.51}+1$

On solving,

  $f(-2.51)=-0.24$

For f(-2.52) substituting the values,

  $f(-2.52)=3{(-2.52)}^2+8(-2.52)-2{(4)}^{-2.52}+1$

On solving,

  $f(-2.52)=-0.05$

For f(-2.53) substituting the values,

  $f(-2.53)=3{(-2.53)}^2+8(-2.53)-2{(4)}^{-2.53}+1$

On solving,

  $f(-2.53)=-0.097$

For f(-2.54) substituting the values,

  $f(-2.54)=3{(-2.54)}^2+8(-2.54)-2{(4)}^{-2.54}+1$

On solving,

  $f(-2.54)=-67.61$

Because, f(2.53) is the closest to 0

Therefore, $x\approx -2.53$

For y, substituting the values,

  $y=3{(-2.53)}^2+8(-2.53)$

On solving,

  $y\approx 1.04$

Hence, the solution is $x\approx -2.53,0.19$ and $y\approx 1.04,1.6$