X²+6x-(a²+2a-8)=0
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Sia ? 6 years, 1 month ago
The given equation is x2+6x−(a2+2a−8)=0
Comparing it with Ax2+Bx+C=0, we get
A=1, B=6 and C=−(a2+2a−8)
∴ D=B2−4AC= (6)2 - 4(1)(-(a2 + 2a - 8))
= 36 +4a2 + 8a - 32
= 4a2 + 8a + 4 = 4(a2 + 2a + 1) = 4(a + 1)2 > 0
So, the given equation has real roots, given by
α=−B+√D2A=−6+√4(a+1)22×1=−6+2(a+1)2 =−3+(a+1)=a−2
β=−B−√D2A=−6−√4(a+1)22×1=−6−2(a+1)2 =−3−(a+1)=−a−4=−(a+4)
Hence, (a - 2) and -(a + 4) are the roots of the given equation.
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