Factorize this sum. Basically I don't even know if this sum is given right or wrong bcz this was a sum of my exam question paper.
x²-x-(a+1)(a+2)
Answers
Step-by-step explanation:
TNPSC Group 1 Exam Eligibility Details: Tamil Nadu Public Service Commission (TNPSC)’s Group I Exam Eligibility details are given below…..
Eligibility Conditions:
I. Age Limit:
Name of the Post Minimum Age
(should have
completed) Maximum Age (should not have completed)
SC’s, SC(A)’s, ST’s,
MBC’s/DC’s, BC’s,
BCM’s and DW’s of
all castes “Others” i.e., candidates
not belonging to SC’s,
SC(A)’s, ST’s, MBC’s/DC’s,
BC’s and BCM’s
1. For all the posts except the
post of Assistant
Commissioner (C.T). 21 Years 37 Years 32 Years
2. Assistant Commissioner (C.T.)
(i) For candidates
possessing any degree 21 Years 37 Years 32 Years
(ii) For candidates possessing
B.L. degree of any
University or Institution
recognized by the
University Grants Commission. 21 Years 38 Years 33 Years
Note:
(i) “Others” (i.e.,) candidates not belonging to SC’s, SC(A’)s, ST’s, MBC’s/DC’s, BC’s and BCM’s who have put in 5 years and more of service in the State / Central Government are not eligible to apply even if they are within the age limit.
II. Educational Qualification: Candidates should possess the following or its equivalent qualification:-
1. Candidates should possess a Degree of any of the Universities incorporated by an Act of the Central or State Legislature in India or any other Educational Institutions established by an Act of Parliament or declared to be deemed as a University under Section 3 of the University Grants Commission Act, 1956.
Note:
(i).The qualification prescribed for these posts should have been obtained by passing the required qualification in the order of studies 10th + HSC (or its equivalent)+ U.G. degree.
(ii). The candidates, who have written the final year degree examination, may also apply for the Preliminary Examination for direct recruitment to the posts included in Group-I Services.
2. Preference will be given to the candidates, who possess the qualification mentioned against each of the following posts:
Sl.
No Name of the Post Qualification
1. Deputy Superintendent of
Police (Category – I) A Degree or Diploma in Criminology and Forensic Science
and also to those who possess National Awards for Physical
Efficiency.
2. Assistant Commissioner
(C.T.) First Preference _ A Degree both in Commerce
and Law together with a Diploma in Taxation laws.
Second Preference _ A Degree both in Commerce
and Law.
Third Preference _ A Degree either in Commerce
or Law together with a Diploma in Taxation laws.
Fourth Preference _ A Degree either in Commerce
or Law.
3. Assistant Director of Rural Development Department Post Graduate Degree in Rural Service of the Gandhigram Rural Institute, Madurai District, Post Graduate Degree or Diploma (Extension, Sociology.
Physical Qualification:
Candidates applying for the post of Deputy Superintendent of Police (Category-I) and District Officer (Fire and Rescue Service) must possess the following Physical qualifications:
Name of the Post Physical Qualification
Deputy Superintendent of
Police (Category-I) i. For Men: Must be not less than 165 cms in Height and not less than 86 cms round the chest on full inspiration and must have a chest expansion of not less than 5 cms on full inspiration.ii. For Women: Must be not less than 155 cms in Height. The Chest measurement will not apply to
them.
District Officer (Fire and Rescue Service) i. For Men: Must be not less than 165 cms in Height, 50 kgs in Weight and not less than 89 cms round the chest on full inspiration and 84 cms (normal) with not less than 5 cms chest expansion on full inspiration.ii. For Women: Must be not less than 155 cms in Height. The Chest measurement and 50 kg Weight will not apply to them
Given x-x−(a+1)(a+2)=0, Here, a = 1, b = 1, c = -(a + 2)(a + 1)
x=
2a
−b±
b
2
−4ac
=
2(1)
−1±
1
2
−4{(a+2)(a+1)}
=
2
−1±
1−4{−(a
2
+3a+2)
x=
2
−1±
1+4a
2
+12a+8
=
2
−1±
4a
2
+12a+9
=
2
−1±
(2a+3)
2
=
2
−1±(2a+3)
∴x=
2
−1+2a+3
and x=
2
−1−2a−3
∴x=
2
2a+2
and x=
2
−2a−4
x=a+1x=−(a+2)
∴ The roots of x
2
+x−(a+2)(a+1)=0 are (a+1) and −(a+2)
hope this helped you Pls mark my answer as brainliest answer