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If the equation $(a+1)x^2-(a+2)x+(a+3)=0$ has roots of equal magnitude and opposite signs, then the roots are

(A) $\pm a$
(B) $\pm\large\frac{1}{2}$$a$
(C) $\pm\large\frac{3}{2}$$a$
(D) $2a$

1 Answer

Toolbox:
  • The sum of the roots of $ax^2+bx+c=0$ is $\large-\frac{b}{a}$
  • The product of the roots is $\large\frac{c}{a}$
Let the roots of $(a+1)x^2-(a+2)x+(a+3)=0$ be $\alpha\:and\:-\alpha$
(Given that the roots are of equal magnitude and opp. signs.)
$\Rightarrow\:\alpha-\alpha=0=\large\frac{a+2}{a+1}$
$\Rightarrow\:a=-2$
$-\alpha^2=\large\frac{a+3}{a+1}$$=\large\frac{1}{-1}$$=-1$
$\Rightarrow\:\alpha=\pm1$$=\pm\large\frac{1}{2}$$a$
answered Jul 21, 2013 by rvidyagovindarajan_1
edited May 21, 2014 by rohanmaheshwari0831_1
Where is the relationship defined between a and alpha? How can we say
±1⇒α=±1=±1=±1/2a
 

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