Problem 54 - Algebra for College Students

Question

For each equation find the value of \(k\) given that 3 satisfies the equation. a) \(x^{4}-3 x^{3}+5 x^{2}-7 x+k=0\) b) \(x^{4}-x^{3}-2 x^{2}+k x-k=0\) c) \(5 x^{3}-k x^{2}-k x-3 k=0\)

Step-by-Step Solution

Answer

a) k = -24b) k = -18c) k = 9

Step 1: Identify given equation and substitute value

For the given equation, substitute the value 3 for the variable x since it's stated that 3 satisfies the equation.

Step 2: Solve for equation a

For equation (a) \(x^{4}-3 x^{3}+5 x^{2}-7 x+k=0\).Substitute x = 3:\[3^{4} - 3(3^{3}) + 5(3^{2}) - 7(3) + k = 0\]Calculate each term:\[81 - 81 + 45 - 21 + k = 0\]Combine like terms:\[24 + k = 0\]Solve for k:\[k = -24\]

Step 3: Solve for equation b

For equation (b)\(x^{4}-x^{3}-2 x^{2}+k x-k=0\).Substitute x = 3:\[3^{4} - 3^{3} - 2(3^{2}) + k(3) - k = 0\]Calculate each term:\[81 - 27 - 18 + 3k - k = 0\]Combine like terms:\[36 + 2k = 0\]Solve for k:\[k = -18\]

Step 4: Solve for equation c

For equation (c)\(5 x^{3}-k x^{2}-k x-3 k=0\).Substitute x = 3:\[5(3^{3}) - k(3^{2}) - k(3) - 3k = 0\]Calculate each term:\[135 - 9k - 3k - 3k = 0\]Combine like terms:\[135 - 15k = 0\]Solve for k:\[k = 9\]

Key Concepts

roots of equations

Roots of equations are the values of the variable that satisfy the equation. In the given exercise, you know that 3 is a root of each equation. This means when you substitute 3 for the variable x, the equation should equal zero. Identifying the roots is crucial because they tell us the specific values that make the polynomial zero. If you substitute a root into the polynomial and it equals zero, this confirms that the value is a true root.

substitution method

The substitution method involves replacing the variable in an equation with a given number. In this case, you need to substitute x with 3 because you are told that 3 satisfies the equation. Here’s how it works step-by-step:

• Take the polynomial equation given.
• Substitute 3 for all occurrences of x in the equation.
• Simplify the resulting expression using arithmetic operations.
• Set the simplified equation equal to zero and solve for the unknown variable (in these exercises, it’s k).

This method helps simplify the equation so you can solve for the unknown variable precisely and effectively.

algebraic manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations to solve for unknown variables. In these solutions, you perform several steps:

• After substitution, calculate powers and products.
• Simplify each term to get a cleaner numeric equation.
• Combine like terms to make the equation manageable.

For instance, in one of the exercises, substituting 3 into the equation might give you \[ 81 - 81 + 45 - 21 + k = 0 \]. You then combine terms to simplify it to \[ 24 + k = 0 \], making it easier to solve for k. Algebraic manipulation is all about making the equation easier to understand and solve.

solving for variables

Solving for variables involves finding the value of an unknown in an equation. After substitution and combining like terms:
\begin{itemize}

You isolate the variable on one side of the equation.

This often means moving terms to the other side to keep the equation balanced.

Divide or multiply as necessary to solve for the variable.

For the given exercises, you isolate k after simplifying. For example, with \[ 24 + k = 0 \], solve for k by subtracting 24 from both sides, giving \[ k = -24 \]. Similarly, in another example, \[ 135 - 15k = 0 \], you divide both sides by 15 to find \[ k = 9 \]. These steps ensure that you find the correct value of the unknown variable.