Question

$\alpha$ and $\beta$ are roots of the quadratic equation $a{x}^2+bx+c=0$. The equation has real roots which are of opposite signs, then the equation $\alpha (x-\beta {)}^2+\beta (x-\alpha {)}^2=0$

• A. has no real roots
• B. has two distinct real roots
• C. has real equal roots
• D. has real roots which are opposite in signs

Answer 1

Answer: (B), (D)

has real roots which are opposite in signs

$\because$Roots are real and distinct
$\therefore D>0$

$\Rightarrow b^2-4ac>0$ & $\alpha \beta <0$

$\alpha \beta =\frac{c}{a}$

$\Rightarrow \frac{c}{a}<0\Rightarrow ac<0$

$\alpha +\beta =-\frac{b}{a}$

$\Rightarrow \alpha (x-\beta {)}^2+\beta (x-\alpha {)}^2=0$
$\Rightarrow (\alpha +\beta )x^2-4\alpha \beta x+\alpha {\beta }^2+{\alpha }^2\beta =0$
$\Rightarrow (\alpha +\beta )x^2-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0$

$\Rightarrow -\frac{b}{a}{x}^2-4\frac{c}{a}x+\frac{c}{a}(-\frac{b}{a})=0$

$\Rightarrow ab{x}^2+4acx+bc=0$

$D=16a^2{c}^2-4a{b}^{2}c=4ac(4ac-b^2)$
$\because 4ac<0\&4ac-b^2<0$

$\therefore D>0$

$\Rightarrow ab{x}^2+4acx+bc=0$

Let ${\alpha }^{{}^{\prime }}.{\beta }^{{}^{\prime }}$ are roots
$\Rightarrow {\alpha }^{{}^{\prime }}.{\beta }^{{}^{\prime }}=\frac{bc}{ab}=\frac{c}{a}<0$

$\Rightarrow {\alpha }^{{}^{\prime }}.{\beta }^{{}^{\prime }}<0$
have two real roots of opposite signs.