Question

How do you solve \[\left( {x + 5} \right)\left( {3x + 1} \right) = 0\]?

Answer 1

Hint: Here in this question, we have to solve the two terms for the variable x. this can be solve by two methods first method is equating each term to the zero and simplify the term for the variable x and secondly, we use the arithmetic operation that is multiplication and then we simplify using the quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], to get the required solution.

Complete step by step solution:
Consider the given terms
\[ \Rightarrow \,\,\left( {x + 5} \right)\left( {3x + 1} \right) = 0\]
Method 1:
We can easily solve this equation by recognizing that the product of two terms is equal to zero if either one of those terms (or both, that's not the case here) is equal to zero.
i.e., the product of \[\left( {x + 5} \right)\] and \[\left( {3x + 1} \right)\] will be equal to zero when
\[ \Rightarrow \,\,\left( {x + 5} \right) = 0\] or \[\left( {3x + 1} \right) = 0\]
\[ \Rightarrow \,\,x + 5 = 0\] \[ 3x + 1 = 0\]
\[ \Rightarrow \,\,x = - 5\] \[ ,x = \dfrac{{ - 1}}{3}\]
Method 2:
Now let us consider the two terms and they are \[\left( {x + 5} \right)\], and \[\left( {3x + 1} \right)\]
Now we have to multiply the terms, to multiply the two terms we use multiplication. The multiplication is one of the arithmetic operations.
Now we multiply the above 2 terms we get
\[\left( {x + 5} \right) \cdot \left( {3x + 1} \right)\]
Here dot represents the multiplication. First, we multiply the first two terms of the above equation
\[ \Rightarrow \,\,x\left( {3x + 1} \right) + 5\left( {3x + 1} \right) = 0\]
On multiplying we get
\[ \Rightarrow \,\,3{x^2} + x + 15x + 5 = 0\]
On simplification we have
\[ \Rightarrow \,\,3{x^2} + 16x + 5 = 0\]
Now, use the quadratic formula to solve for the solutions.
i.e., \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here, \[a = 3\], \[b = 16\], and \[c = 5\],then
\[ \Rightarrow \,\,x = \dfrac{{ - 16 \pm \sqrt {{{16}^2} - 4\left( 3 \right)\left( 5 \right)} }}{{2\left( 3 \right)}}\]
\[ \Rightarrow \,\,x = \dfrac{{ - 16 \pm \sqrt {256 - 60} }}{6}\]
\[ \Rightarrow \,\,x = \dfrac{{ - 16 \pm \sqrt {196} }}{6}\]
As, we know the 196 is the square number of 14, i.e., \[\sqrt {196} = \sqrt {{{14}^2}} = 14\], then
\[ \Rightarrow \,\,x = \dfrac{{ - 16 \pm 14}}{6}\]
\[ \Rightarrow \,\,x = \dfrac{{ - 16 + 14}}{6}\] or \[ \Rightarrow \,\,x = \dfrac{{ - 16 - 14}}{6}\]
\[ \Rightarrow \,\,\,x = \dfrac{{ - 2}}{6}\] or \[ \Rightarrow \,\,x = \dfrac{{ - 30}}{6}\]
On simplification, we get
\[ \Rightarrow \,\,x = - \dfrac{1}{3}\] or \[ \Rightarrow \,\,x = - 5\]
Hence, we get the same solution in both method
The required solution is \[x = - \dfrac{1}{3}\] or \[x = - 5\]

Note: To multiply we use operation multiplication, multiplication of numbers is different from the multiplication of algebraic expression. In the algebraic expression it involves the both number that is constant and variables. Variables are also multiplied when multiplying mainly, we have to concentrate on sign conversion, if the variable is the same then the result will be in the form of exponent.