Chapter 9, Problem 34SGR

To determine

To solve the given equation by factoring.

Answer to Problem 34SGR

The value of k is $-\frac{1}{3}$and 3 .

Explanation of Solution

Given information :

The equation provided is $3k^2-8k=3$ . This can be written as $3k^2-8k-3=0$ .

Formula used :

The given equation can be solved by factoring. To do so, break up the midpoint to arrive at two factors. Then use the zero-product property to equate each factor in order to get the roots.

Calculation :

  $3k^2-8k-3=0$

Equation to be solved.

  $=>3k^2-9k+k-3=0$

Breaking the middle term that when multiplied, provides that values of $a{x}^2\times c$ .

  $=>3k\left(k-3\right)+1\left(k-3\right)=0$

Arranging the common factors.

  $=>\left(3k+1\right)\left(k-3\right)=0$

The two factors of the function.

Now, the first root is:

  $3k+1=0$

Using zero-product property.

  $k=-\frac{1}{3}$

The second root is:

  $k=3$

Using zero-product property.

  $k=3$

Thus, the values of k are $-\frac{1}{3}$and 3 .

To check the solutions, put them in the given equation and verify that left-hand and right-hand sides are equal.

For $k=-\frac{1}{3}$

  $3{\left(-\frac{1}{3}\right)}^2-8\left(-\frac{1}{3}\right)-3=0$

Putting the value in the equation.

  $=>\frac{1}{3}+\frac{8}{3}-3=0$

Simplifying the expression.

  $=>\frac{8}{3}-\frac{8}{3}=0$

Hence this result is a success.

Now, for $k=3$

  $3{\left(3\right)}^2-8\left(3\right)-3=0$

Putting the value in the equation.

  $=>27-24-3=0$

Simplifying the expression.

  $=>27-27=0$

Hence this result is a success.