# : The solution set of $ (ax + b)(cx - d) = 0 $ is

(A) $ \left\{ {\dfrac{b}{a},\dfrac{d}{c}} \right\} $

(B) $ \left\{ {a,c} \right\} $

(C) $ \left\{ { - \dfrac{b}{a},\dfrac{d}{c}} \right\} $

(D) $ \left\{ {b, - d} \right\} $

Answer

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**Hint :**Firstly let us get some knowledge about the solution set. The set of all variables that makes the equation valid is called a solution set. Plug in each value from the replacement set and compare both sides of the equation to find the solution set from the replacement set. The equation is valid if the two sides are identical, and the value is a solution.

**Complete Step By Step Answer:**

Let's get some more knowledge about the solution set. Any value of a variable that renders the stated equation valid is referred to as a solution. The set of all variables that makes the equation valid is referred to as a solution set. Since\[2\left( 4 \right){\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}14\], the solution set for \[2y{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}14\]is\[4\]. Since \[{2^2}{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}5\left( 2 \right)\]and\[{3^2}{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}5y\], the solution set for \[{y^2}{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}5y\]is (2, 3). If an equation has no solutions, the solution set is the empty set or null set, which is a set with no members and is denoted by the symbol $ \emptyset $ . The solution set to \[{x^2}{\text{ }} = {\text{ }} - {\text{ }}9\]is, for example, that no number squared equals a negative number.

Now,

$ (ax + b)(cx - d) = 0 $

Equating the factors individually to zero we will get:

$ (ax + b) = 0 \\

ax = - b \\

x = - \dfrac{b}{a} \\

(cx - d) = 0 \\

cx = d \\

x = \dfrac{d}{c} \\ $

So from above solution we can say that the solution set is: $ \left\{ { - \dfrac{b}{a},\dfrac{d}{c}} \right\} $

**So option(C) is correct.**

**Note :**

The replacement set is a collection of values that can be used to replace a variable. Plug in each value from the replacement set and compare both sides of the equation to find the solution set from the replacement set. The equation is valid if the two sides are identical, and the value is a solution.

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