Question

If the roots of the equation $b{x^2} + cx + a = 0$ be imaginary, then for all real values of x, the expression $3{b^2}{x^2} + 6bcx + 2{c^2}$is
(A). Greater than – 2ab
(B). Less than – 2 ab
(C). Greater than – 4ab
(D). Less than – 4ab

Answer 1

Hint: Before attempting this question, one must have prior knowledge about the concept of quadratic equation and also remember to use discriminant i.e. ${b^2} - 4ac$which explains the nature of roots of the quadratic equation, using this information will help you to approach the solution of the problem.

Complete step-by-step answer:
According to the given information we have a quadratic equation i.e. ${b^2} + cx + a = 0$where roots of the equation are imaginary
Now using the property of discriminant of the quadratic formula according to which discriminant identifies the nature of the roots of the quadratic equation which is given as ${b^2} - 4ac$
Since the roots of the given quadratic equation i.e. ${b^2} + cx + a = 0$ are imaginary in nature
Therefore, ${b^2} - 4ac < 0$
Substituting the value from given quadratic equation i.e. ${b^2} + cx + a = 0$ in the formula of discriminant we get
${c^2} - 4ab < 0$
$ \Rightarrow $${c^2} < 4ab$
$ \Rightarrow $$ - {c^2} > - 4ab$ (equation 1)
Now let’s take $f\left( x \right) = 3{b^2}{x^2} + 6bcx + 2{c^2}$
Since by the above equation we can say that $3{b^2} > 0$
Therefore, we can say that the above equation has minimal value and since the minimum value is calculated as; Minimum value = $\dfrac{{ - D}}{{4a}}$
Substituting the values in the above formula we get
Minimum value = $\dfrac{{4ac - {b^2}}}{{4a}}$
Now substituting the values from the quadratic equation i.e. $3{b^2}{x^2} + 6bcx + 2{c^2}$ we get
Minimum value = $\dfrac{{4\left( {3{b^2}} \right)\left( {2{c^2}} \right) - {{\left( {6bc} \right)}^2}}}{{4\left( {3{b^2}} \right)}}$
$ \Rightarrow $ Minimum value = $\dfrac{{24{b^2}{c^2} - 36{b^2}{c^2}}}{{12{b^2}}}$
$ \Rightarrow $ Minimum value = $\dfrac{{ - 12{b^2}{c^2}}}{{12{b^2}}}$
$ \Rightarrow $ Minimum value = $ - {c^2}$
So, from the equation 1
$ - {c^2} > - 4ab$
Therefore, for all real value of x is greater than – 4ab
Hence, option C is the correct option.

Note: In the above equation we came across the term “quadratic equation” which can be explained as a polynomial equation which consists of degree two. A quadratic equation is represented as $a{x^2} + bx + c$here “a” is the first coefficient of x2, b is the coefficient of x and c is the absolute term in the quadratic equation. A quadratic equation of degree two consist of two roots which satisfies the quadratic equation the roots of quadratic equation can be find by using the quadratic formula i.e. $(a,b) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}$here a and b are the two roots of the quadratic equation.