Chapter 4.1, Problem 56E

To determine

To find:values of B and C.

Answer to Problem 56E

  $\begin{array}{l}B=1\\ C=-2\end{array}$

Explanation of Solution

Given information: Critical Thinking: If B and C are real roots of $x^2+Bx+C=0,$ where $B\ne 0\text{ and }C\ne 0$ .

Calculation:

Since B and C are the roots of the equation

Since B and C are real roots of $x^2+Bx+C=0$

We can write

  $x^2+Bx+C=\left(x-B\right)\left(x-C\right)$

Expand the equation

  $\begin{array}{l}{x}^2+Bx+C=\left(x-B\right)\left(x-C\right)\\ x^2+Bx+C=x^2-Cx-Bx+BC\end{array}$

Group like terms

  $x^2+Bx+C=x^2-\left(B+C\right)x+BC$

Equate both sides

  $\begin{array}{l}BC=C\\ B+C=-B\end{array}$

Solve for B

  $\begin{array}{l}BC=C\\ B=\frac{C}{C}\\ B=1\end{array}$

Solve for C

  $\begin{array}{l}B+C=-B\\ C=-B-B=-2B\end{array}$

Substitute B value

  $C=-2\ast 1=-2$

So the equation becomes $x^2+Bx+C=0$ and

  $\begin{array}{l}B=1\\ C=-2\end{array}$