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Both the roots of the equation `(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0`are alwaysa. positive b. realc. negative d. noneof theseA. positiveB. negativeC. realD. real and equal |
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Answer» Correct Answer - C `(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0` `rArr 3x^(2) - 2 (a + b + c)x + (ab + bc + ca) = 0` Now `D = 4 (a + b+ c)^(2) - 12(ab + bc + ca)` `= 4(a^(2) + b^(2) + c^(2) - ab - bc - ca)` `= 2[(a - b)^(2) + (b - c)^(2) + (c - a)^(2)]` which is always positive or zero so roots are real |
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