(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$. \\ Let $t = x-a$. Then the equation becomes: \\ $t(t-1) + (t-1)(t-2) + t(t-2) = 0$ \

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(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$. \\ Let $t = x-a$. Then the equation becomes: \\ $t(t-1) + (t-1)(t-2) + t(t-2) = 0$ \\ $t^2 - t + t^2 - 3t + 2 + t^2 - 2t = 0$ \\ $3t^2 - 6t + 2 = 0$ \\ The discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$. \\ Since $D > 0$,the roots for $t$ are real and distinct. \\ Consequently,$x = a + t$ will also have real and distinct roots.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$,where $a \in R$,are always:

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A
equal
B
imaginary
C
real and distinct
D
rational and equal

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$,where $a \in R$,are always:

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