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प्रश्न
Find the nature of the roots of the equation 4x2 – 4a2x + a4 – b4 = 0, b ≠ 0.
बेरीज
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उत्तर
4x2 – 4a2x + a4 – b4 = 0
Here a = 4, b = – 4a2, c = a4 – b4
D = b2 – 4ас
⇒ (– 4a2)2 – 4 × 4 × (a4 – b4)
⇒ 16a4 – 16(a4 – b4)
= 16a4 – 16a4 + 16b4
= 16b4 > 0
D > 0
Hence, roots are real and unequal.
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