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Maths-

General

Easy

Question

If the roots of the quadratic equation x² + 6x + b = 0 are real and distinct and they differ by atmost 4 then the least value of b is –

5
6
7
8

hint Hint:

In the question, it is given that the roots of the quadratic equation $x squared plus 6 x plus b space equals space 0$ are real and distinct and they differ by atmost 4. We are required to find the least value of b.

The correct answer is: 5

Step by step solution:
Given equation: $x squared plus 6 x plus b space equals space 0$
It is given that the roots of the equation are real and distinct.
Therefore, $B squared minus 4 A C space greater than space 0 space$
$rightwards double arrow$ $6 squared minus 4 left parenthesis 1 right parenthesis left parenthesis b right parenthesis space greater than space 0$
$rightwards double arrow$ $36 space minus space 4 b greater than 0$
$rightwards double arrow 4 b less than 36$
$rightwards double arrow b less than 9$
It is also given that the roots differ by atmost 4
Therefore, $open vertical bar alpha minus beta close vertical bar less or equal than 4$
$rightwards double arrow open vertical bar alpha minus beta close vertical bar squared less or equal than 16$
$rightwards double arrow open parentheses alpha plus beta close parentheses squared minus 4 alpha beta less or equal than 16$
$rightwards double arrow open parentheses negative 6 close parentheses squared minus 4 left parenthesis b right parenthesis less or equal than 16$ [ since sum of roots = -6 and product of roots = b]
$rightwards double arrow 36 minus 4 b less or equal than 16$
$rightwards double arrow 4 b greater or equal than 20$
$rightwards double arrow b greater or equal than 5$
Thus, least value of b is 5
Hence, option(b) is the correct option.

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