Question

How do you find the zeros of $3{x^2} + 3x + 1$ ?

Answer 1

Hint: The given equation is in the form of a quadratic equation. We will use the quadratic equation formula to find the zero of the given equation. A polynomial having degree 2 in one variable in form of $f\left( x \right) = a{x^2} + bx + c$ where $a,b,c \in R$ and $a \ne 0$, when equated to zero becomes quadratic equation of degree 2 . The general form of the quadratic equation is $f\left( x \right) = a{x^2} + bx + c$. $a$ is called as the leading coefficient and $c$ is called absolute term of quadratic equation. The value satisfying the quadratic equation is called the roots of the quadratic equation. The quadratic equation will always have two roots. The nature of the root may be either real or imaginary.
Formula for finding the root of the quadratic equation is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
The rule of signs for quadratic polynomial states that the number of positive real roots of ${P_n}\left( x \right) = 0$ will not be more than the number of sign changes and the number of negative roots will not be more than the number of sign changes in polynomial ${P_n}\left( { - x} \right) = 0$.

Complete step by step solution:
Step: 1 the given quadratic equation is $3{x^2} + 3x + 1$.
Compare the given polynomial with the general form of quadratic equation $f\left( x \right) = a{x^2} + bx + c$.
Therefore, $a = 3$ $b = 3$ and $c = 1$.
Step: 2 Substitute the value of $a,b,$ and $c$ in the formula.
$
   \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{{\left( 3 \right)}^2} - 4 \times 3 \times 1} }}{{2 \times 3}} \\
   \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 - 12} }}{{2 \times 3}} \\
 $
Solve the equation to find the value of $x$. Substitute the value of ${i^2} = - 1$ in the equation.
$
   \Rightarrow x = \dfrac{{3 \pm i\sqrt 3 }}{6} \\
   \Rightarrow x = \dfrac{{3 \pm i\sqrt 3 }}{6} \\
 $
Step: 3 consider the positive value of $x$.
 $
   \Rightarrow x = \dfrac{{3 + i\sqrt 3 }}{6} \\
   \Rightarrow x = \dfrac{{3 + i\sqrt 3 }}{6} \\
 $
Consider the negative value of $x$.
$
   \Rightarrow x = \dfrac{{3 - i\sqrt 3 }}{6} \\
   \Rightarrow x = \dfrac{{3 - i\sqrt 3 }}{3} \\
 $

Final Answer:
Therefore the roots of the quadratic equation are $x = \dfrac{{3 - i\sqrt 3 }}{6}$ and $x = \dfrac{{3 + i\sqrt 3 }}{6}$.

Note:
While solving the problems students are advised to use the formula of quadratic equation $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Compare the value of given quadratic equation with general equation of quadratic equation $f\left( x \right) = a{x^2} + bx + c$ to find the coefficient of leading term and absolute term.