Question

Let a and b be nonzero real roots of the quadratic equation x²+ ax + b = 0 and a + b, a -b and - a -b be the roots of the equation x⁴+ax³+cx²+dx+e=0. Then which of the following statement is false?

• A. a=0
• B. c=0
• C. d=0
• D. e=0

Answer 1

The correct option is B c=0

$x^2+ax+b=0$
$\alpha +\beta =-a,\alpha \beta =b$
$x^4+a{x}^3+c{x}^2+dx+e=0$
Roots are $\alpha +\beta ,\alpha -\beta ,-\alpha +\beta$ and $-\alpha -\beta$
Sum of roots $=-a$
$\alpha +\beta +\alpha -\beta -\alpha +\beta -\alpha -\beta =-a$
$a=0\Rightarrow \alpha +\beta =0$
Sum of roots taken two at a time $=+c$ $(\alpha +\beta )(\alpha -\beta )+(\alpha -\beta )(-\alpha +\beta )+(-\alpha +\beta )(-\alpha -\beta )+(-\alpha -\beta )(\alpha +\beta )+(\alpha +\beta )(-\alpha +\beta )+(\alpha -\beta )(-\alpha +\beta )=c$
0- $(\alpha +\beta {)}^2+0+0+0-(\alpha +\beta {)}^2=c$
$-2(\alpha -\beta {)}^2=c$
$-2[{\alpha }^2+{\beta }^2-2\alpha \beta ]=c$
$-2\left[\right(\alpha +\beta {)}^2-2\alpha \beta -2\alpha \beta ]=c$
$-2[-4\alpha \beta ]=c$
$c=8\alpha \beta$
$c=8b\ne 0$
Product of roots taken three at a time $=-d$ $(\alpha +\beta )(\alpha -\beta )(-\alpha +\beta )+(\alpha +\beta )(-\alpha +\beta )(-\alpha -\beta )+(-\alpha -\beta )(\alpha +\beta )+(\alpha +\beta )(-\alpha +\beta )(-\alpha +\beta )(-\alpha +\beta )=-d$
$0+0+0+0=-d$
$d=0$
Product of roots $=e$ $(\alpha +\beta )(\alpha -\beta )(-\alpha +\beta )(-\alpha -\beta )=e$
$e=0$
So, $c=0$ is the false statement