Question

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

Answer 1

Hint: In this question, we have to find the value \[x\]. We will use the fact that, If \[a \times b = 0\], this implies either \[a = 0\] , \[b = 0\] or both are zero. We will be equated with each of the equations \[(x + 5)\] and \[(3x + 1)\] to \[0\]. Then we will simplify to find the value of \[x\]. This value of \[x\] is the required solution of the given equation.

Complete step by step answer:
This question is based on a quadratic equation. Quadratic equation is of the form \[a{x^2} + bx + c = 0\], where \[a\], \[b\] and \[c\] are real constants. The quadratic equation can be factored into a product of linear equations. Each of them, when equated to zero, gives the solution of the quadratic equation.

Consider the given question,
We are given, \[(x + 5)(3x + 1) = 0\]
We know that, if \[a \times b = 0\], this implies either \[a = 0\] , \[b = 0\] or both are zero.
Hence, we have
\[ \Rightarrow (x + 5) = 0\]
Adding \[( - 5)\] to both sides, we have
\[ \Rightarrow x + 5 - 5 = - 5\]
On simplifying, we get
\[ \Rightarrow x = - 5\]
And on solving other equation, we have
\[ \Rightarrow 3x + 1 = 0\]
Adding \[( - 1)\] to both sides, we get
\[ \Rightarrow 3x + 1 - 1 = - 1\]
On solving, we get
\[ \Rightarrow 3x = - 1\]
Dividing both side by \[3\], we get
\[ \therefore x = \dfrac{{ - 1}}{3}\]
Hence \[x = - 5\] or \[x = \dfrac{{ - 1}}{3}\]

Therefore, the solution of \[(x + 5)(3x + 1) = 0\] is either \[x = - 5\] or \[x = \dfrac{{ - 1}}{3}\].

Note: Alternatively, we can solve as follow, we can write the given equation in form of quadratic equation, as
\[ \Rightarrow (x + 5)(3x + 1) = 0\]
On multiplying each term of \[(x + 5)\] by each term of \[(3x + 1)\], we have
\[ \Rightarrow 3{x^2} + 15x + x + 5 = 0\]
On simplifying we have
\[ \Rightarrow 3{x^2} + 16x + 5 = 0\]
Now comparing with the standard equation, we have,
\[a = 3\], \[b = 16\] and \[c = 5\]
Hence, the solution is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
i.e., \[x = \dfrac{{ - 16 \pm \sqrt {{{16}^2} - 4 \times 3 \times 5} }}{{2 \times 3}}\]
on solving, we have
\[x = \dfrac{{ - 16 \pm \sqrt {256 - 60} }}{6} = \dfrac{{ - 16 \pm \sqrt {196} }}{6} = \dfrac{{ - 16 \pm 14}}{6}\]
0n solving, we have
Hence, \[x = \dfrac{{ - 1}}{3}\] and \[x = - 5\].