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Algebra Level 4

If unity is double repeated root of $px^3 + q(x^2 + x) + r = 0$ then

None of these choices

pqr > 0

pr < 0

pq < 0

pqr < 0

1 solution

Raghav Rathi
Jan 4, 2016

$\quad It\quad is\quad given\quad that\quad \\ \quad \quad \quad p{ x }^{ 3 }+q({ x }^{ 2 }+x)+r\quad has\quad 1\quad as\quad twice\quad repeated\quad root.\\ .....\\ so,\quad let\quad the\quad expression\quad above\quad be\quad f(x).\\ \quad \quad \quad \quad \quad f'(x)\quad =\quad 3p{ x }^{ 2 }+2qx+q.\\ ..\quad f'(1)=0\quad \star \{ \because ,\quad f(x)\quad has\quad 1\quad as\quad twice\quad repeated\quad root.\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Hence\quad 1\quad is\quad only\quad root\quad of\quad f'(x)\} \\ \rightarrow \quad 3p+3q=0\quad \Rightarrow \quad p\quad =\quad -q\\ \quad \\ \quad Hence,\quad \boxed { pq<0 } \\ \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$

How to prove that $f'(x)$ will have $1$ as a root? @Raghav Rathi

Anik Mandal - 5 years, 3 months ago

The graph of the equation will touch the x axis at 1(condition for repeated roots), hence extrema will also be at 1.

Raghav Rathi - 5 years, 3 months ago

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