Question

if $a,\,b,\,c \in R$ and $a + b + c = 0$, then the quadratic equation $4a{x^2} + 3bx + 2c = 0$ has
A) One positive and one negative
B) Imaginary roots
C) Real roots
D) None of the above

Answer 1

Hint:
To justify which type of roots is a quadratic equation, it is justified by the discriminant, with the help of the discriminant roots are classified. Find the discriminant and justify the root type.

Complete step by step solution:
Given, $a,\,b,\,c \in R$ and $a + b + c = 0.......\left( 1 \right)$
For a quadratic equation $p{x^2} + qx + r = 0$, the discriminant $\Delta = {q^2} - 4pr........\left( 2 \right)$
From the quadratic equation $4a{x^2} + 3bx + 2c = 0$ find the discriminant.
With the help of (2), we have
$
   \Rightarrow \Delta = {q^2} - 4pr \\
   \Rightarrow \Delta = {\left( {3b} \right)^2} - 4\left( {4a} \right)\left( {2c} \right) \\
   \Rightarrow \Delta = 9{b^2} - 32ac........\left( 3 \right) \\
 $
From (1), we have $a + b + c = 0$, substitute $b = - \left( {a + c} \right)$ in (3).
$
   \Rightarrow \Delta = 9{b^2} - 32ac \\
   \Rightarrow \Delta = 9{\left( { - \left( {a + c} \right)} \right)^2} - 32ac \\
   \Rightarrow \Delta = 9\left( {{a^2} + {c^2} + 2ac} \right) - 32ac \\
   \Rightarrow \Delta = 9{a^2} + 9{c^2} + 18ac - 32ac \\
   \Rightarrow \Delta = 9{a^2} + 9{c^2} - 14ac \\
   \Rightarrow \Delta = 7{a^2} + 7{c^2} - 14ac + 2{a^2} + 2{c^2} \\
   \Rightarrow \Delta = 7\left( {{a^2} + {c^2} - 2ac} \right) + 2\left( {{a^2} + {c^2}} \right) \\
   \Rightarrow \Delta = 7{\left( {a - c} \right)^2} + 2\left( {{a^2} + {c^2}} \right) \\
 $
Here, ${\left( {a - c} \right)^2}\,$ and ${a^2} + {c^2}$ are greater than 0, as a, c are real numbers.

Hence, the roots of the equation $4a{x^2} + 3bx + 2c = 0$ are real roots.

Note:
To justify which type of root discriminant is the way to find out which type of root it is. Real numbers can be in fractions, positive, or negative. we know that $\sqrt{-1}=i$. So, Any root with a negative sign inside the root will be complex.