SATHEE: Theory of Equations 1 Question 50

Theory of Equations 1 Question 50

51. For $a \leq 0$ , determine all real roots of the equation

$x^{2} - 2 a | x - a | - 3 a^{2} = 0$

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Answer:

Correct Answer: 51. $(x = a (1 - \sqrt{2}), a (\sqrt{6} - 1))$

Solution:

Here, $a \leq 0$

Given, $x^{2} - 2 a | x - a | - 3 a^{2} = 0$

Case I When $x \geq a$

$\begin{aligned} \Rightarrow & x^{2} - 2 a (x - a) - 3 a^{2} = 0 \\ \Rightarrow & x^{2} - 2 a x - a^{2} = 0 \\ \Rightarrow & x = a \pm \sqrt{2} a \end{aligned}$

[as $a (1 + \sqrt{2}) < a$ and $a (1 - \sqrt{2}) > a$ ]

$∴$ Neglecting $x = a (1 + \sqrt{2})$ as $x \geq a$

$\Rightarrow x = a (1 - \sqrt{2})$

Case II When $x < a \Rightarrow x^{2} + 2 a (x - a) - 3 a^{2} = 0$

$\Rightarrow x^{2} + 2 a x - 5 a^{2} = 0 \Rightarrow x = - a \pm \sqrt{6} a$

$[as a (\sqrt{6} - 1) < a and a (- 1 - \sqrt{6}) > a]$

$∴$ Neglecting $x = a (- 1 - \sqrt{6}) \Rightarrow x = a (\sqrt{6} - 1)$

From Eqs. (i) and (ii), we get

$x = a (1 - \sqrt{2}), a (\sqrt{6} - 1)$

Theory of Equations 1 Question 50

51. For $a \leq 0$ , determine all real roots of the equation

Answer:

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Theory of Equations 1 Question 50

51. For a≤0, determine all real roots of the equation

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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51. For $a \leq 0$ , determine all real roots of the equation