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प्रश्न
Solve: x2 + 6x – (a2 + 2a – 8) = 0.
बेरीज
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उत्तर
x2 + 6x – (a2 + 2a – 8) = 0
⇒ x2 + 6x – (a + 4)(a – 2) = 0
⇒ x2 + [(a + 4) – (a – 2)]x – (a + 4)(a – 2) = 0
⇒ x2 + (a + 4)x – (a – 2)x – (a + 4)(a – 2) = 0
⇒ x[x + (a + 4)] – (a – 2)[x + (a + 4)] = 0
⇒ [x + (a + 4)][x – (a – 2)] = 0
⇒ x + (a + 4) = 0 or x – (a – 2) = 0
⇒ x = –(a + 4) or x = (a – 2)
Hence, –(a + 4) and (a – 2) are the roots of the given equation.
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