A quadratic equation is given as $x^2+2(a-1)x+(a-5)=0$ then what could be the possible value of a if:

a) The equation has positive roots

b) The equation has roots of opposite sign

c) The equation has negative roots

I tried to check the nature of discriminant but it takes me nowhere.Can someone tell me the best procedure to deal with such sums?Or hints atleast..

  • 1
    $\begingroup$ Hint: the sum of the roots is $2(a-1)$ and their product is $a-5$. $\endgroup$ – Cla Dec 15 '15 at 14:03

The equation $f(x)=0$ with $f(x)=x^2+2(a-1)x+(a-5)$ has

  1. (a) two positive roots, iff $f(0)>0$ and $f'(0)<0$
  2. (b) roots of opposite sign, iff $f(0)<0$
  3. (c) two negative roots, iff $f(0)>0$ and $f'(0)>0$

This can be seen graphically, since $f$ is a quadratic polynomial with positive sign for $x^2$



Use the fact that, if $x_1$ and $x_2$ are the solutions than $$ x_1+x_2=-2(a-1) \quad \land \quad x_1x_2=a-5 $$ so:

$x_1=-x_2 \Rightarrow a-1=0$

$x_1>0$ and $x_2>0$ $\Rightarrow x_1x_2>0$ and $x_1+x_2>0$

and... can you do from this?


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