Chapter 2, Problem 1P

An analysis of a circuit shown in Fig. P2.1 yields the quadratic equation for the current I as $3I^2-6I=45$, where I is in amps.

(a) Rewrite the above equation in the form $I^2+aI+b=0$, where a and b are constants.

(b) Solve the equation in part (a) by each of the following methods: factoring, compacting the square, and the quadratic formula.

To determine

(a)

To rewrite the given current relation in the form of $I^2+aI+b=0$.

Answer to Problem 1P

The required equation of current in the form of $I^2+aI+b=0$ is

  $I^2+2I+\left(-15\right)=0$

Explanation of Solution

Given:

The current relation is given as

  $3I^2+6I=45$. ...... (1)

Calculation:

Divide by $3$ on both sides in the equation (1)

  $I^2+2I=15$

Rearrange above equation.

  $I^2+2I-15=0$

The required equation of current in the form of $I^2+aI+b=0$ is

  $I^2+2I+\left(-15\right)=0$

Conclusion:

Thus, the required equation of current in the form of $I^2+aI+b=0$ is

  $I^2+2I+\left(-15\right)=0$

To determine

(b)

Solve the equation $I^2+2I-15=0$ by factoring, completing the square, and the quadratic formula.

Answer to Problem 1P

The equation is solved by factoring, completing the square, and the quadratic and the roots of the equation are $3$ and $-5$.

Explanation of Solution

Concept Used:

The roots of the quadratic equation can be determined by the quadratic formula given as-

  $I=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ...... (2)

Calculation:

Substitute $2$ for $b$, $1$ for $a$ and $-15$ for $c$ in equation (2)

  $\begin{array}{c}I=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(-15\right)}}{2\left(1\right)}\\ =\frac{-2\pm 8}{2}\end{array}$

This equation gives two values of current

  $I=3\text{ amps}$ and $I=-5\text{ amps}$

For completing square method,

Rearrange equation (1)

  $I^2+2I=15$

Add $1$ on both side of the above equation.

  $\begin{array}{c}{I}^2+2I+1=15+1\\ {\left(I+1\right)}^2=16\end{array}$

Rearrange the above equation.

  $\begin{array}{c}{\left(I+1\right)}^2-16=0\\ {\left(I+1\right)}^2-{\left(4\right)}^2=0\\ \left(I+1-4\right)\left(I+1+4\right)=0\\ \left(I-3\right)\left(I+5\right)=0\end{array}$

The value of current $I$ are $3\text{ amps}$ and $-5\text{ amps}$

For factoring method,

Rearrange equation (1)

  $\begin{array}{c}{I}^2+\left(5-3\right)I-15=0\\ I^2+5I-3I-15=0\\ I\left(I+5\right)-3\left(I+5\right)=0\\ \left(I+5\right)\left(I-3\right)=0\end{array}$

The value of current $I$ are $3\text{ amps}$ and $-5\text{ amps}$