Question

LU, 12, 24, 26, 24, 22, 22, 28, 26, 22.
Find the mean and mode of the following tabulation:
Age (in years)
12
13
14
15
16
17
Number of Students
1
3
9
12
8
7
Height (in cm)
112
114
116
118
120
Number of Girls
3
6.
10​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.