Question

How do you solve \[4{x^2} + 12x + 3 = 0\] ?

Answer 1

Hint: From the given question, we can see that the equation is a quadratic equation. When an equation is in the form of \[a{x^2} + bx + c = 0\] , it is a quadratic equation. Here we will use the quadratic formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to solve our question.

Complete step-by-step answer:
According to the question, the given equation is:
 \[4{x^2} + 12x + 3 = 0\]
We know that the above equation is in the form of a quadratic equation. The form of the quadratic equation is:
 \[a{x^2} + bx + c = 0\]
When we relate it with our given equation, we get:
 \[a = 4\,;\,b = 12\,;\,c = 3\]
Now, we will use the quadratic formula to solve the equation, the formula is:
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where a, b and c are constants.
By putting the values of a, b and c in the equation, we get:
  \[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {{{(12)}^2} - 4 \times 4 \times 3} }}{{2 \times 4}}\]
Now, we will try to solve the equation here, and we will get:
 \[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {144 - 48} }}{8}\]
 \[ \Rightarrow x = \dfrac{{13 \pm \sqrt {96} }}{8}\]
We know that 96 is not a square root of any whole number. So, we will try to simplify the root, and we will get:
 \[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {4 \times 4 \times 6} }}{8}\]
 \[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {{4^2} \times 6} }}{8}\]
 \[ \Rightarrow x = \dfrac{{ - 12 \pm 4\sqrt 6 }}{8}\]
Now, we will take out the common factor from here. The common factor here is 4:
 \[ \Rightarrow x = \dfrac{{4( - 3 \pm \sqrt 6 )}}{8}\]
Here, we can cancel or divide 4 from 8, and we will get:
 \[ \Rightarrow x = \dfrac{{( - 3 \pm \sqrt 6 )}}{2}\]
Now, we will solve for ‘x’. We will get two values of ‘x’ because it is a quadratic equation:
 \[x = \dfrac{{ - 3 + \sqrt 6 }}{2}\] and \[x = \dfrac{{ - 3 - \sqrt 6 }}{2}\]
Therefore, after solving, we get the values of ‘x’ as \[x = \dfrac{{ - 3 + \sqrt 6 }}{2}\,\,and\,\,\dfrac{{ - 3 - \sqrt 6 }}{2}\,\] .
So, the correct answer is “\[x = \dfrac{{ - 3 + \sqrt 6 }}{2}\,\,and\,\,\dfrac{{ - 3 - \sqrt 6 }}{2}\,\] ”.

Note: We can solve the given quadratic equation also by completing square method, in this we have to complete the square in the given equation with help of arithmetic operations as follows
We have given \[4{x^2} + 12x + 3 = 0\]
From the equation,
 \[ \Rightarrow 4{x^2} + 12x + 3 = 0\]
We can write it as
 \[ \Rightarrow {\left( {2x} \right)^2} + 2 \times 6x + 3 = 0\]
Now, to complete the square, adding ${3^2}$ and also to maintain the balance of the equation, subtracting ${3^2}$ too, we will get
 \[
   \Rightarrow {\left( {2x} \right)^2} + 2 \times 2x \times 3 + {3^2} + 3 - {3^2} = 0 \\
   \Rightarrow {\left( {2x + 3} \right)^2} + 3 - 9 = 0 \\
   \Rightarrow {\left( {2x + 3} \right)^2} - 6 = 0 \\
   \Rightarrow {\left( {2x + 3} \right)^2} = 6 \;
 \]
Taking square root both sides, we will get
 \[
   \Rightarrow 2x + 3 = \pm \sqrt 6 \\
   \Rightarrow 2x = - 3 \pm \sqrt 6 \\
   \Rightarrow x = \dfrac{{ - 3 \pm \sqrt 6 }}{2} \;
 \]
So we again get the answer from this method.