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Find the number of real root of the equa...

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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The correct Answer is:
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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