(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 $t(t-1)+(t-1)(t-2)+t(t-2)=0$.Expanding the t

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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 $t(t-1)+(t-1)(t-2)+t(t-2)=0$.Expanding the terms: $(t^2-t) + (t^2-3t+2) + (t^2-2t) = 0$.Combining like terms: $3t^2 - 6t + 2 = 0$.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 be real and distinct for any $a \in R$.

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 $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$ for $a \in R$ are always:

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

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The product of real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$ is:

If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

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