Question

Solve this by factorisation method only

Answer 1

Answer:

Hello friend !!

x2+5x−(a2+a−6)=0

⇒x2+5x−(a2+3a−2a−6)=0

⇒x2+5x−(a(a+3)−2(a+3))=0

⇒x2+5x−(a+3)(a−2)=0

⇒x2+[(a+3)−(a−2)]x−(a+3)(a−2)=0

⇒x2+(a+3)x−(a−2)x−(a+3)(a−2)=0

⇒x[x+(a+3)]−(a−2)[x+(a+3)]=0

⇒[x+(a+3)][x−(a−2)]=0

⇒[x+(a+3)]=0 or [x−(a−2)]=0

⇒x=−(a+3) or x=(a−2)

Hope it helps.

Answer 2

Step-by-step explanation:

X2+5x _ (a2+a_6)=0

= X2+5x_( a2 + 3a _ 2a _6)=0

= X2+5x _ [ a ( a+3) _2 ( a+3)] =0

= x 2+5× _ ( a+3) (a_2) =0

= x2+[ ( a+3) _ ( a_2) ] × _ ( a+3) ( a_2) =0

= x2 + (a+3) ×_( a_2) x _ (a+3) (a_3) =0

= x [ x+(a+3)] _ (a_2) [ x +(a+3) ] =0

= [ x+( a+3 )] [ x _(a_2)] =0

= [x+(a+3)] =9 or [x_( a_2)] =0

= x = _ ( a+3) or x = ( a_2)

make me brainliest.