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11 May 2022, 02:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
80% (01:10) correct
20%
(01:16)
wrong
based on 75
sessions
History
Date
Time
Result
Not Attempted Yet
18 May 2022, 00:12
Kudos
Bookmarks
CASE-I
BY SOLVING QUAD EQ, WE GET X=-2 AND X=1. SO NO UNIQUE VALUE. HENCE AD CAN BE ELEMINATED
CASE-II, AGAIN SOLVING QUAD EQ, WE GET X=-3 AND X=1. AGAIN NOTSUFFICIENT SINCE NO UNIQUE VALUE IS THERE
BUT WHEN WE COMBINE BOTH CASES, WE GET COMMON X WHICH IS EQUAL TO 1.
SO MY ANSWERL WILL BE C
PLEASE CORRECT ME, IF I AM WRONG.
BY SOLVING QUAD EQ, WE GET X=-2 AND X=1. SO NO UNIQUE VALUE. HENCE AD CAN BE ELEMINATED
CASE-II, AGAIN SOLVING QUAD EQ, WE GET X=-3 AND X=1. AGAIN NOTSUFFICIENT SINCE NO UNIQUE VALUE IS THERE
BUT WHEN WE COMBINE BOTH CASES, WE GET COMMON X WHICH IS EQUAL TO 1.
SO MY ANSWERL WILL BE C
PLEASE CORRECT ME, IF I AM WRONG.
15 Jul 2022, 08:01
Kudos
Bookmarks
Quadratic equation solved by sum and product of the roots.
Attachments
E3236F8A-086C-42A1-8686-568B2E92A1EC.jpeg [ 457.4 KiB | Viewed 1123 times ]
