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: (A)

p=q≠r

Answer 2

Let the two equations be: (1) $px^2+qx+r=0$ (2) $x^3-3x^2+3x+2=0$

Rewrite equation (2) as $(x-1)^3 + 3 = 0$, which gives $(x-1)^3 = -3$. Let $y = x-1$. Then $y^3 = -3$. The roots for $y$ are $k, k\omega, k\omega^2$, where $k$ is a cube root of $-3$ and $\omega$ is a complex cube root of unity. The roots for $x$ are $1+k, 1+k\omega, 1+k\omega^2$. Since the coefficients of the quadratic equation are real, its two common roots with the cubic equation must be the complex conjugate pair: $\alpha = 1+k\omega$ and $\beta = 1+k\omega^2$.

Using Vieta's formulas for $px^2+qx+r=0$: Sum of roots: $\alpha + \beta = (1+k\omega) + (1+k\omega^2) = 2 + k(\omega+\omega^2) = 2 + k(-1) = 2-k$. So, $-q/p = 2-k$.

Product of roots: $\alpha\beta = (1+k\omega)(1+k\omega^2) = 1 + k(\omega+\omega^2) + k^2\omega^3 = 1 + k(-1) + k^2(1) = 1-k+k^2$. So, $r/p = 1-k+k^2$.

From $-q/p = 2-k$, we get $k = 2+q/p$. Substitute this into $r/p = 1-k+k^2$: $r/p = 1 - (2+q/p) + (2+q/p)^2$ $r/p = -1 - q/p + 4 + 4q/p + q^2/p^2$ $r/p = 3 + 3q/p + q^2/p^2$ Multiply by $p^2$: $rp = 3p^2 + 3pq + q^2$.

Check option Ⓐ ($p=q \neq r$): If $p=q$, the equation becomes $rp = 3p^2 + 3p(p) + p^2 = 7p^2$. Since $p \neq 0$, we can divide by $p$: $r = 7p$. This means $p=q$ and $r=7p$. If $p \neq 0$, then $p \neq r$. Thus, $p=q \neq r$ is consistent.