Question

Show that equation (a - 2)x^2 + (2 - b)x + (b - a) = 0 has equal roots, if 2a = b + 2.

Plz Ans this​

Answer 1

Answer: a=8

Step-by-step explanation:

If 2a =b+2

=>b=2a-2

Putting the value of b in equation ,

i.e, ( a-2)x^2 +(2-b)x +(b-a) =0

=> (a-2)x^2+{2-(2a-2)}x +{(2a-2)-a} =0

=> (a-2)x^2 +(2-2a+2)x +(2a-2-a)=0

=> (a-2)x^2 +(4-2a)x +(a-2)=0

As we know discriminant ,D = b^2-4ac

And for the equal roots, D=0

Expressing as D ,

b^2 -4ac = (4-2a)^2- { 4 (a-2)(a-2) }=0

=> (16+4a^2-16a) - {4 (a-2)^2} =0

=> (16+4a^2-16a)-4 (a^2 +4 -4a)=0

=> 16+4a^2 -16a -4a^2 -16 +16a=0

=> 4a^2 -32a =0

=> 4a- 32 =0

=> a=32/4

=>a= 8

Hope it helps ...............