Statement-1 : If roots of the equation $x^2-b x+c=0$ are two consecutive integers then $b^2-4 c=1$ \

Question

Statement-1 : If roots of the equation $x^2-b x+c=0$ are two consecutive integers then $b^2-4 c=1$ \Statement-2: If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are odd integer, then the roots of the equation $4 a b c x^2+\left(b^2-4 a cight) x-b=0$ are real and distinct

Professor PC Thomas Classes · Answer

$x^{2}-bx+c=0$. $|\alpha-\beta|=1 \Rightarrow (\alpha+\beta)^{2}-4\alpha\beta=1 \Rightarrow b^{2}-4c=1$. Statement 2: $4abcx^{2}+(b^{2}-4ac)x-b=0 \Rightarrow D=(b^{2}-4ac)^{2}+16ab^{2}c = (b^{2}+4ac)^{2} > 0$. Roots are real and distinct.