Question

Consider the following statements:
$S_1:$ if the roots fo $x^2-bx+c=0$ are two consecutive integers, then value of $b^2-4c$ is equal to $1$
$S_2:$ If $\alpha ,\beta$ are roots of $x^2-x+3=0$ then value of ${\alpha }^4+{\beta }^4$ is equal $7$.
$S_3:$ If $\alpha ,\beta ,\gamma$ are roots of $x^3-7x^2+16x-12=0$ then value of ${\alpha }^2+{\beta }^2+{\gamma }^2$ is equal $17$.
State, in order whether $S_1,S_2,S_2$ are true or false.

• A. True
• B. False

Answer 1

The correct option is A True

$S_1:$ If the roots of $x^2-bx+c=0$ are two consecutive integers

i.e., $|\alpha -\beta |=\left|\frac{\sqrt{b^2-4ac}}{a}\right|=1$

$⟹\left|\frac{\sqrt{(-b{)}^2-4c}}{1}\right|=1$

$⟹\sqrt{b^2-4c}=1$

$\therefore b^2-4c=1$

$S_2:\alpha ,\beta$ are the roots of $x^2-x+3=0$

$\alpha +\beta =\frac{-b}{a}=1$ and $\alpha \beta =\frac{c}{a}=3$

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

$={\left[\right(\alpha +\beta {)}^2-2\alpha \beta ]}^2-2(\alpha \beta {)}^2$

$={\left[\right(1{)}^2-2\left(3\right)]}^2-2(3{)}^2$

$=(1-6{)}^2-2(9)$

$=25-18$

$=7$

$S_3:\alpha ,\beta ,\gamma$ are the roots of $x^3-7x^2+16x-12=0$

$\alpha +\beta +\gamma =\frac{-b}{a}=7$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=16$

and $\alpha \beta \gamma =\frac{-d}{a}=12$

${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$

$=7^2-2\left(16\right)$

$=49-32$

$=17$

So, $S_1,S_2,S_3$ all are true (Option A)