# - Question-If the equation $a{x^2} + 2bx - 3c = 0$ has non real roots and (3c/4)<(a+b); then C is always

A. <0

B. >0

C.$ \geqslant $ 0

D. Zero

Answer

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Hint-Make use of the formula of the discriminant when the roots are non real(imaginary) and solve it.

Given that the roots of the equation are non real

The equation given is $a{x^2} + 2bx - 3c = 0$

On comparing with the standard form $a{x^2} + bx + c = 0$ ,we can write the value of a=a, b=2b,c=-3c

Now, we know that the value of the discriminant when the roots are non real is less than zero

So, we have ${b^2} - 4ac < 0$

On substituting the values of a, b, c we can write

\[\begin{gathered}

{\left( {2b} \right)^2} - 4a( - 3c) < 0 \\

\Rightarrow 4{b^2} + 12ac < 0 \\

\Rightarrow 4{b^2} < - 12ac, \\

\Rightarrow {b^2} < - 3ac, \\

\Rightarrow - a > {b^2}/(3c),a < - {b^2}/(3c) \\

\end{gathered} \]

Also we have in the equation

$4{b^2} + 12ac < 0$

$4{b^2}$ will always be positive

So, this implies to say that $12ac < 0$

$ \Rightarrow c < 0{\text{ or a < 0}}$

Also given in the question that (3c/4)<(a+b) ----(i)

We already had the value which said $a < - {b^2}/3c$

So, let’s put that value of a in eq(i)

So ,we get 3c/4<$( - {b^2}/3c + b)$---(ii) (We are putting the maximum value of a here)

So, this above equation implies that the value of c<0

So, option A is the correct answer.

Note: In equation (ii) we need not solve the entire equation, by inspection we can easily

make out that the value of c<0 also make sure to use the correct inequality of discriminant in accordance to the nature of the root given.

Given that the roots of the equation are non real

The equation given is $a{x^2} + 2bx - 3c = 0$

On comparing with the standard form $a{x^2} + bx + c = 0$ ,we can write the value of a=a, b=2b,c=-3c

Now, we know that the value of the discriminant when the roots are non real is less than zero

So, we have ${b^2} - 4ac < 0$

On substituting the values of a, b, c we can write

\[\begin{gathered}

{\left( {2b} \right)^2} - 4a( - 3c) < 0 \\

\Rightarrow 4{b^2} + 12ac < 0 \\

\Rightarrow 4{b^2} < - 12ac, \\

\Rightarrow {b^2} < - 3ac, \\

\Rightarrow - a > {b^2}/(3c),a < - {b^2}/(3c) \\

\end{gathered} \]

Also we have in the equation

$4{b^2} + 12ac < 0$

$4{b^2}$ will always be positive

So, this implies to say that $12ac < 0$

$ \Rightarrow c < 0{\text{ or a < 0}}$

Also given in the question that (3c/4)<(a+b) ----(i)

We already had the value which said $a < - {b^2}/3c$

So, let’s put that value of a in eq(i)

So ,we get 3c/4<$( - {b^2}/3c + b)$---(ii) (We are putting the maximum value of a here)

So, this above equation implies that the value of c<0

So, option A is the correct answer.

Note: In equation (ii) we need not solve the entire equation, by inspection we can easily

make out that the value of c<0 also make sure to use the correct inequality of discriminant in accordance to the nature of the root given.

Last updated date: 21st Sep 2023

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