(D) Given $z^2 + z + 1 - a = 0$. Using the quadratic formula,$z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}$. Since the

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(D) Given $z^2 + z + 1 - a = 0$. Using the quadratic formula,$z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}$. Since the imaginary part of $z$ is non-zero,the discriminant must be negative. Thus,$4a - 3 Among the given options,$\frac{3}{4}$ is not less than $\frac{3}{4}$,so $a$ cannot take the value $\frac{3}{4}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^2 + z + 1$ is real. Then $a$ cannot take the value

IIT 2012 All IIT

A
$-1$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$\frac{3}{4}$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^2 + z + 1$ is real. Then $a$ cannot take the value

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