Multiple choice

The equation ${ x }^{ 2 } - ax + b = 0 & { x }^{ 3 } - p{ x }^{ 2 } + qx = 0$, where $b \neq 0, q \neq 0$, have one common root & the second equation has two equal roots. Then, $2\left( q + b \right) = $.

  1. $2ap$
  2. $ap^{2}$
  3. $a^2p$
  4. $ap$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the common root be k. k^2 - ak + b = 0 and k^3 - pk^2 + qk = 0. The second equation has two equal roots, say r, r, and s. So x^3 - (2r+s)x^2 + (r^2+2rs)x - r^2s = 0. Comparing coefficients, p = 2r+s, q = r^2+2rs. The common root k must be r. Substituting k=r into the first equation: r^2 - ar + b = 0, so b = ar - r^2. Then 2(q+b) = 2(r^2+2rs + ar - r^2) = 2(2rs + ar) = 2r(2s+a). Through substitution, this simplifies to ap.

AI explanation

The equation x^3 - px^2 + qx = 0 factors as x(x^2 - px + q) = 0. Since q is not 0, the common root must be from the quadratic factor, which has two equal roots, making its discriminant p^2 - 4q = 0 and its double root equal to p/2. Because this root is also common to x^2 - ax + b = 0, we substitute it to get (p/2)^2 - a(p/2) + b = 0. Multiplying by 4 yields p^2 - 2ap + 4b = 0, and substituting p^2 with 4q results in 4q - 2ap + 4b = 0, which simplifies to 2(q + b) = ap.

Keep practicing — related questions