Reduce x²+7x+12/x²-6x-27
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Given: x
2
+6x−7=0
x
2
+6x+9−9−7=0
x
2
+3x+3x+9=+16
x(x+3)+3(x+3)=16
(x+3)(x+3)=16
(x+3)
2
=±(4)
2
∴x+3=±4
2×a×b=6x
2×x×b=6x(∵a=x)
b=
2x
6x
b=3
b
2
=9
Taking square root on both the sides,
(Half of coefficient of x =
2
6
∴b=3)
If, x + 3 = 4 x +3 = - 4
x = 4 - 3 x = - 4 - 3
x = 1 x = - 7
∴ 1 and - 7 are the roots of x
2
+6x−7=0
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