Question

17.
If (4a + c)2 < 4b2 then one root of ax2 + bx + c = 0 lies in
(1) (-2,2)
(2) (-1, 1)
(3) (-
$\infty$
-2)
(4) (2,
$\infty$


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Answer 1

Given info : if (4a + c)² < 4b² , then one of the roots of ax² + bx + c lies in

Solution : here, (4a + c)² < 4b²

⇒(4a + c)² < (2b)²

⇒(4a + c)² - (2b)² < 0

⇒(4a + c - 2b)(4a + c + 2b) < 0

(4a + c - 2b) < 0 or, 4a + c + 2b > 0 ......(1)

Now, here we see equation is ax² + bx + c = 0

If we assume, one of the roots lies between -2 to 2. It means value of equation at x = -2 will be negative while at x = 2 will be positive

i.e., a(-2)² + b(-2) + c < 0 or, a(2)² + b(2) + c > 0

⇒4a - 2b + c < 0 or, 4a + 2b + c > 0 these are what we get above eq (1).

Hence one of the roots of given equation lies in (-2, 2). i.e., option (1) is correct.