Multiple choice

The equations $x^2-px+6=0$ and $x^2+2x-r=0$ one root as $3$, then value of $p$ and $r$ are

  1. $3,6$
  2. $5,15$
  3. $2,3$
  4. $3,-5$
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B Correct answer
Explanation

Since 3 is a root of x^2 - px + 6 = 0, substitute x=3 to get 9 - 3p + 6 = 0, which gives 3p = 15, so p = 5. Since 3 is a root of x^2 + 2x - r = 0, substitute x=3 to get 9 + 6 - r = 0, which gives r = 15.

AI explanation

Since 3 is a root of both equations, we substitute 3 for x in x^2 - px + 6 = 0 to get 9 - 3p + 6 = 0, which gives 3p = 15 and p = 5. Next, substitute 3 for x in x^2 + 2x - r = 0 to get 9 + 6 - r = 0, yielding r = 15. Therefore, the values are p = 5 and r = 15.

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