Question

If a, b, c \[\in \] R and abc < 0, then the equation \[bc{{x}^{2}}+2\left( b+c-a \right)x+a=0\] has
a)Both positive roots
b)Both negative roots
c)Real roots
d)One positive and one negative root

Answer 1

Hint: To solve the question, we have to apply the formula of the product of the roots of the quadratic equation for analysing the signs of the roots of the given equation. To find the nature of the roots, apply the discriminant formula of the quadratic equation to the given equation.
Complete step-by-step answer:
The given equation is \[bc{{x}^{2}}+2\left( b+c-a \right)x+a=0\]
We know that for a general quadratic equation \[p{{x}^{2}}+qx+r=0\],
The sum of roots of the equation is given by \[\dfrac{-q}{p}\]
The product of roots of the equation is given by \[\dfrac{r}{p}\]
On comparing the general quadratic equation with the given quadratic equation, we get
The value of p = bc, q = 2(b + c - a), r = a . BY substituting these values in the above given formula, we get
The sum of roots of the given equation is equal to \[\dfrac{-2(b+c-a)}{bc}\]
The product of roots of the given equation is equal to \[\dfrac{a}{bc}\]
Given that abc < 0, which can be possible only if either of a, b, c is negative considering the other two values are positive or all the values of a, b, c are negative.
Thus, we get the product of roots of the given equation is negative.
\[\dfrac{a}{bc}<0\]
Thus, the roots of the equation are of opposite signs.
We know the discriminant is positive for real roots. The formula for discriminant of general quadratic equation is given by \[{{q}^{2}}-4pr\]. By substituting the values, we get
\[\begin{align}
  & {{\left( 2(b+c-a) \right)}^{2}}-4\left( bc \right)(a) \\
 & =4{{\left( b+c-a \right)}^{2}}-4\left( abc \right) \\
 & =4\left( {{\left( b+c-a \right)}^{2}}-abc \right)>0 \\
\end{align}\]
Since abc < 0, the whole expression is positive.
Thus, the roots of the given quadratic equation are real.
Hence, option (c) and (d) are the right choices.
Note: The possibility of mistake can be, not using the product of the roots of the quadratic equation for analysing the nature and signs of the roots of the given equation. The alternative way of solving the question is by substituting some real values for a, b, c such that the condition of abc < 0 is satisfied. Thus, we can arrive at the nature of roots of the given quadratic equation.