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4-2.Quadratic Equations and Inequations
easy
यदि $a < 0$ तब असमिका $a{x^2} - 2x + 4 > 0$ के मूल निम्न द्वारा प्रदर्शित होंगे
A
$\frac{{1 + \sqrt {1 - 4a} }}{a} > x > \frac{{1 - \sqrt {1 - 4a} }}{a}$
B
$x < \frac{{1 - \sqrt {1 - 4a} }}{a}$
C
$x < 2$
D
$2 > x > \frac{{1 + \sqrt {1 - 4a} }}{a}$
Solution
$a{x^2} – 2x + 4 > 0$
$\Rightarrow$ $x = \frac{{2 \pm \sqrt {4 – 16a} }}{{2a}}$
$⇒x = \frac{{1 \pm \sqrt {1 – 4a} }}{a}$
$\therefore$ $\frac{{1 – \sqrt {1 – 4a} }}{a} < x < \frac{{1 + \sqrt {1 – 4a} }}{a}$.
Standard 11
Mathematics
