factorise:ax²+(1+a²)x+a
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Answered by
1
Answer:
a {x}^{2} + (1 + {a}^{2} )x + aax
2
+(1+a
2
)x+a
To find :-
It's factorization
Solution:-
a {x}^{2} + (1 + {a}^{2} )x + aax
2
+(1+a
2
)x+a
a {x}^{2} + x + {a}^{2} x + a = 0ax
2
+x+a
2
x+a=0
a {x}^{2} + {a}^{2} x + x + a = 0ax
2
+a
2
x+x+a=0
Taking ax and 1 as common,
We get,
ax(x + a) + 1(x + a) = 0ax(x+a)+1(x+a)=0
(ax + 1)(x + a) = 0(ax+1)(x+a)=0
(ax + 1) = 0 \: \:(ax+1)=0
(x + a) = 0(x+a)=0
ax + 1 = 0ax+1=0
x = \dfrac{ - 1}{a}x=
a
−1
or,
x = - ax=−a
hence, the solution is
x = \dfrac{ - 1}{a} \: \: \: \: x = - ax=
a
−1
x=−a
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1
Answer:
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