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Find the number of real root of the equation`|0x-a x-b x+a0x-c x+b x+c0|=0,a!=b!=ca n db(a+c)> a c` |
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Answer» Correct Answer - three Expanding Using Sarrus Rule `Delta =(x-a)(x+b)(x-c)+(x-b)(x+a)(x+c)` `=2x(x^(2)+ac-ab-bc)` Now `Delta=0` gives x=0 or `x^(2) =b(a+c)-ac` if b(a+c) `gt ac` we have three roots `0+- sqrt({b(a+c)-ac})` |
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