Question

(c) Expansion of (2a + 3b + 4C + d)^2is
(0) 4a2 + 9b2 + 1602 + d2 + ab + 8cd + 160c + 24bc + 4ad + obd
4a? + 9b2 + 160? + d + 12ab + 8cd + Bac + 12bc + 2 ad + 3bd
4a2 + 9b? + 160? + d2 + 12ab + 8cd + 1600 + 24bc + 4ad + obd
(iv) 4a2 + 9b? + 160° + d + 12ab + 8cd + 32ac + 48bc + Sad + 12bd​

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.