Mark the tick against the correct answer in the following:
The solution set of the equation is
To find: Value of x
We have,
Applying
⇒
Applying
⇒
Taking 2 common from R1
⇒
Taking 2 common from R2
⇒
Applying
⇒
Expanding along R1
⇒ 4[x{(x)(a+x) – (-x)(a-2x)}] – (-x){(0)(a+x) – (-x)(a)}] = 0
⇒ 4[x{ax + x2 +ax -2x2}] – (-x){ax}] = 0
⇒ 4[x{2ax - x2}] + ax2] = 0
⇒ 4[2ax2 – x3 + ax2] = 0
⇒ – x2 + 3ax = 0
⇒ -x(x – 3a) = 0
⇒ x = 0 , or x = 3a