Question

The equ $px^2+qx+r=0$ and $x^3-3x^2+3x+2=0$ have two roots in common then

• A. p=q≠r
• B. p≠q=r
• C. p≠q≠r
• D. p=q=r

Answer 1

Answer: (C)

p≠q≠r

Answer 2

The cubic equation is $x^3-3x^2+3x+2=0$. This can be written as $(x-1)^3 + 3 = 0$, so $(x-1)^3 = -3$. The roots are $x_1 = 1 - \sqrt[3]{3}$, $x_2 = 1 - \sqrt[3]{3}\omega$, and $x_3 = 1 - \sqrt[3]{3}\omega^2$, where $\omega$ is a complex cube root of unity. The quadratic $px^2+qx+r=0$ has two roots in common. Assuming real coefficients for $p, q, r$, these common roots must be a complex conjugate pair. Thus, the roots are $x_2$ and $x_3$. The sum of these roots is $x_2+x_3 = 2 - \sqrt[3]{3}(\omega+\omega^2) = 2 + \sqrt[3]{3}$. The product of these roots is $x_2x_3 = 1 - \sqrt[3]{3}(\omega+\omega^2) + 3 = 4 + \sqrt[3]{3}$. The quadratic equation is proportional to $x^2 - (2+\sqrt[3]{3})x + (4+\sqrt[3]{3}) = 0$. So, $p=k$, $q=-k(2+\sqrt[3]{3})$, $r=k(4+\sqrt[3]{3})$ for some $k \neq 0$. Since $\sqrt[3]{3}$ is irrational, $2+\sqrt[3]{3} \neq -1$, $4+\sqrt[3]{3} \neq 1$, and $-(2+\sqrt[3]{3}) \neq 4+\sqrt[3]{3}$. Therefore, $p \neq q$, $p \neq r$, and $q \neq r$, which means $p \neq q \neq r$.