Chapter 4.5, Problem 57E

(a)

To determine

To find: the value of $k$ for given equation so that the given $x$ value is a solution.

Answer to Problem 57E

The value of $k=60$ .

Explanation of Solution

Given information:

Given equation,

  $x^3-6x^2-7x+k=0;x=4$

Calculation:

Because $x=4$ is a solution of equation, so substitute $x=4$ in equation and solve for $k$ .

  $\begin{array}{c}{\left(4\right)}^3-6{\left(4\right)}^2+k=0\\ 64-96-28+k=0\\ k=60\end{array}$

Therefore, the value of $k=60$ .

(b)

To determine

To find: the value of $k$ for given equation so that the given $x$ value is a solution.

Answer to Problem 57E

The value of $k=33$ .

Explanation of Solution

Given information:

Given equation,

  $2x^3+7x^2-kx-18=0;x=-6$

Calculation:

Because $x=-6$ is a solution of equation, so substitute $x=-6$ in equation and solve for $k$ .

  $\begin{array}{c}2{\left(-6\right)}^3+7{\left(-6\right)}^2-k\left(-6\right)-18=0\\ -432+252+6k-18=0\\ 6k=198\\ k=33\end{array}$

Therefore, the value of $k=33$ .

(c)

To determine

To find: the value of $k$ for given equation so that the given $x$ value is a solution.

Answer to Problem 57E

The value of $k=6$ .

Explanation of Solution

Given information:

Given equation,

  $k{x}^3-35x^2+19x+30=0;x=5$

Calculation:

Because $x=5$ is a solution of equation, so substitute $x=5$ in equation and solve for $k$ .

  $\begin{array}{c}k{\left(5\right)}^3-35{\left(5\right)}^2+19\left(35\right)+30=0\\ 125k-875+665+30=0\\ 125k=750\\ k=6\end{array}$

Therefore, the value of $k=6$ .