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Question: If the roots of the equation \(b{x^2} + cx + a = 0\) be imaginary, then for all real values of x, th...

If the roots of the equation bx2+cx+a=0b{x^2} + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2+6bcx+2c23{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

Explanation

Solution

Before attempting this question, one must have prior knowledge about the concept of quadratic equation and also remember to use discriminant i.e. b2−4ac{b^2} - 4acwhich 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. b2+cx+a=0{b^2} + cx + a = 0where 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 b2−4ac{b^2} - 4ac
Since the roots of the given quadratic equation i.e. b2+cx+a=0{b^2} + cx + a = 0 are imaginary in nature
Therefore, b2−4ac<0{b^2} - 4ac < 0
Substituting the value from given quadratic equation i.e. b2+cx+a=0{b^2} + cx + a = 0 in the formula of discriminant we get
c2−4ab<0{c^2} - 4ab < 0
⇒\Rightarrow c2<4ab{c^2} < 4ab
⇒\Rightarrow −c2>−4ab - {c^2} > - 4ab (equation 1)
Now let’s take f(x)=3b2x2+6bcx+2c2f\left( x \right) = 3{b^2}{x^2} + 6bcx + 2{c^2}
Since by the above equation we can say that 3b2>03{b^2} > 0
Therefore, we can say that the above equation has minimal value and since the minimum value is calculated as; Minimum value = −D4a\dfrac{{ - D}}{{4a}}
Substituting the values in the above formula we get
Minimum value = 4ac−b24a\dfrac{{4ac - {b^2}}}{{4a}}
Now substituting the values from the quadratic equation i.e. 3b2x2+6bcx+2c23{b^2}{x^2} + 6bcx + 2{c^2} we get
Minimum value = 4(3b2)(2c2)−(6bc)24(3b2)\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 = 24b2c2−36b2c212b2\dfrac{{24{b^2}{c^2} - 36{b^2}{c^2}}}{{12{b^2}}}
⇒\Rightarrow Minimum value = −12b2c212b2\dfrac{{ - 12{b^2}{c^2}}}{{12{b^2}}}
⇒\Rightarrow Minimum value = −c2 - {c^2}
So, from the equation 1
−c2>−4ab- {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 ax2+bx+ca{x^2} + bx + chere “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)=−b±b2−4ac2ac(a,b) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}here a and b are the two roots of the quadratic equation.