Question

Tlustration (10) (With GST) from the following information of M/s. Makkar and Co., Delhi, prepare the
Purchases Book for the month of February, 2019
2019
Feb 1 Purchased from M/s. Black and Co., Kolkata
5 Gross pencils @ 600 per gross
2 dozen registers @ 250 per dozen
Less Trade Discount @ 10%
Feb 5 Purchased for cash from the Student Mart
20 dozen exercise books @150 per dozen
Feb 10 Purchased from The Paparika Co., Delhi
8 reams of white paper @ 250 per ream
10 reams of ruled paper @ 300 per ream
Less Trade Discount @ 10%
Feb 20 Purchased 80 Reynolds Pens @* 10 each from M/s. Sharma Bros., Delhi
CGST and SGST is levied @ 2.5% each and IGST is levied @ 5%.​

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.