(D) Let $f(x) = a x^{2} + b x + c$. Given that $f(1) = a + b + c < 0$. Since the quadratic equation $a x^{2} + b x + c = 0$ has imaginary roots, the g

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(D) Let $f(x) = a x^{2} + b x + c$. Given that $f(1) = a + b + c Since the quadratic equation $a x^{2} + b x + c = 0$ has imaginary roots, the graph of $f(x)$ does not intersect the $x$-axis. This means $f(x)$ is either always positive $(a > 0)$ or always negative $(a Since $f(1) Also, $f(0) = c$. Since $f(x)$ is always negative, $f(0) Thus, $a < 0$ and $c < 0$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Quadratic expressions and Position of roots

Easy

Let $a, b, c$ be real numbers such that $a+b+c < 0$ and the quadratic equation $a x^{2}+b x+c=0$ has imaginary roots. Then:

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A
$a > 0, c > 0$
B
$a > 0, c < 0$
C
$a < 0, c > 0$
D
$a < 0, c < 0$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Quadratic expressions and Position of roots

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Let $a, b, c$ be real numbers such that $a+b+c < 0$ and the quadratic equation $a x^{2}+b x+c=0$ has imaginary roots. Then:

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If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

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