Solve the following polynomial equation for x : x2 – 2(a + 2)x + (a + 1) (a + 3) = 0.
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1
Answer:
Given x
2
+x−(a+2)(a+1)=0, Here, a = 1, b = 1, c = -(a + 2)(a + 1)
x=
2a
−b±
b
2
−4ac
=
2(1)
−1±
1
2
−4{(a+2)(a+1)}
=
2
−1±
1−4{−(a
2
+3a+2)
x=
2
−1±
1+4a
2
+12a+8
=
2
−1±
4a
2
+12a+9
=
2
−1±
(2a+3)
2
=
2
−1±(2a+3)
∴x=
2
−1+2a+3
and x=
2
−1−2a−3
∴x=
2
2a+2
and x=
2
−2a−4
x=a+1x=−(a+2)
∴ The roots of x
2
+x−(a+2)(a+1)=0 are (a+1) and −(a+2)
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