Question

How do you solve $3{x^2} + 6x + 3 = 0$?

Answer 1

Hint:
According to the question we have to determine the solution of the given quadratic expression which is$3{x^2} + 6x + 3 = 0$. So, first of all to find the solution or we can say that the roots/zeros of the given quadratic expression we have to rearrange the terms of the expression which can be done by dividing the whole expression by 3.
Now, with the help of the coefficient of ${x^2}$ and the constant term we have to determine the coefficient of x by finding the factors of the product of ${x^2}$ and the constant term.
Now, have to take the terms common which can be taken as the common term in the expression obtained.
Now, we have to solve the factors which are as obtained in the solution step 3 to determine the roots/zeroes.

Complete step by step solution:
Step 1: First of all to find the solution or we can say that the roots/zeros of the given quadratic expression we have to rearrange the terms of the expression which can be done by dividing the whole expression by 3. Hence,
$ \Rightarrow \dfrac{{3{x^2} + 6x + 3}}{3} = \dfrac{0}{3}$
On solving the expression as obtained just above,
$ \Rightarrow {x^2} + 2x + 1 = 0$
Step 2: Now, with the help of coefficient of ${x^2}$and the constant term we have to determine the coefficient of x by finding the factors of the product of ${x^2}$and the constant term. Hence,
$
   \Rightarrow {x^2} + (1 + 1)x + 1 = 0 \\
   \Rightarrow {x^2} + x + x + 1 = 0 \\
 $
Step 3: Now, have to take the terms common which can be taken as the common term in the expression obtained. Hence,
$
   \Rightarrow x(x + 1) + 1(x + 1) = 0 \\
   \Rightarrow (x + 1)(x + 1) = 0 \\
 $
Step 4: Now, we have to solve the factors which are as obtained in the solution step 3 to determine the roots/zeroes. Hence,
$
   \Rightarrow (x + 1) = 0 \\
   \Rightarrow x = - 1 \\
 $
And,
$
   \Rightarrow (x + 1) = 0 \\
   \Rightarrow x = - 1 \\
 $

Hence, we have determined the required values of x for the given quadratic expression which is$3{x^2} + 6x + 3 = 0$ are $x = - 1, - 1$.

Note:
1) It is necessary that we have to determine the coefficient of x with the help of the product of the coefficient of ${x^2}$ and the constant term.
2) On solving a quadratic expression only two possible roots/zeroes can be obtained which will satisfy the given quadratic expression mean on placing these in place of x the whole expression becomes 0.