Question

If $x^{2} + \alpha x + \beta = 0$ then $p \neq \alpha$

• A. 4
• B. 6
• C. 3
• D. 2

Answer 1

Answer: (A)

4

Answer 2

We have $\frac{p - q}{r - p},1$

∴$\frac{q - r}{p - q},1$

$\frac{r - p}{p - q},1$

∴$1,\frac{q - r}{p - q}$

$x^{2} + px + 12 = 0$.