Q26 of 110 Page 270

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

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