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प्रश्न
If a, b are real then show that the roots of the equation (a – b)x2 – 6(a + b)x – 9(a – b) = 0 are real and unequal
बेरीज
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उत्तर
(a – b)x2 – 6(a + b)x – 9(a – b) = 0
Δ = b2 – 4ac
= (– 6(a + b)2 – 4(a – b)(– 9(a – b))
= 36(a + b)2 + 36(a – b)2
= 36(a2 + 2ab + b2) + 36(a2 – 2ab + b2)
= 72a2 + 12b2
= 72(a2 + b2) > 0
∴ The roots are real and unequal.
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