x^2-4x-(a^2+2a-3) solve by middle term splitting
Answers
Step-by-step explanation:
x
2
+4x−(a
2
+2a−3)=0
\implies \sf{ {x}^{2} + 4x - ( {a}^{2} + 3a - a- 3) \: } = 0⟹x
2
+4x−(a
2
+3a−a−3)=0
\implies \sf{ {x}^{2} + 4x - \bigg [ {a}(a + 3) - 1(a + 3) \bigg] \: } = 0⟹x
2
+4x−[a(a+3)−1(a+3)]=0
\implies \sf{ {x}^{2} + 4x - (a + 3) (a - 1) \: } = 0⟹x
2
+4x−(a+3)(a−1)=0
\implies \sf{ {x}^{2} + \bigg[ (a + 3) - (a - 1) \bigg] x - (a + 3) (a - 1) \: } = 0⟹x
2
+[(a+3)−(a−1)]x−(a+3)(a−1)=0
\implies \sf{ {x}^{2} + (a + 3) x- (a - 1) x - (a + 3) (a - 1) \: } = 0⟹x
2
+(a+3)x−(a−1)x−(a+3)(a−1)=0
\implies \sf{ {x}(x+ a + 3) - (a - 1) (x + a + 3) \: } = 0⟹x(x+a+3)−(a−1)(x+a+3)=0
\implies \sf{ (x+ a + 3) (x - a + 1)\: } = 0⟹(x+a+3)(x−a+1)=0
Since the product of two real numbers are zero then either of them are zero
\implies \sf{ (x+ a + 3) = 0 \: \: \: or \: \: \: (x - a + 1)\: } = 0⟹(x+a+3)=0or(x−a+1)=0
Now
\sf{ (x+ a + 3) = 0 \:} \: \: \: gives(x+a+3)=0gives
\sf{ x = - ( a + 3) \:} \: \: \:x=−(a+3)
Again
\sf{ (x - a + 1) = 0 \:} \: \: \: gives(x−a+1)=0gives
\sf{ x = a - 1 \:} \: \: \:x=a−1
Hence the required solution is
\sf{x = - (a+3 ) \: \: , \: \: (a-1)}x=−(a+3),(a−1)