Question

The value of p for which roots of the quadratic equation $x^2 - px + 6 = 0$ are rational, is

• A. $1$
• B. $-5$
• C. $25$
• D. $\sqrt{5}$

Hint:

For multiple-choice questions, directly substitute the options into the equation to see which one leads to simple factorization.
Substituting $p = -5$ gives $x^2 + 5x + 6 = 0$, which instantly factors into $(x+2)(x+3) = 0$!

Answer 1

The Correct Option is B

Step 1: Try substituting each option and checking for clean factorisation, instead of computing the discriminant formula first.
For rational roots we want $x^2-px+6=0$ to factor neatly with integers.
Step 2: Test $p=-5$.
The equation becomes $x^2+5x+6=0$, which factors as $(x+2)(x+3)=0$, giving roots $x=-2,-3$. Both are rational.
Step 3: Quickly rule out the others.
$p=1$ gives $x^2-x+6=0$ with discriminant $1-24=-23<0$ (no real roots); $p=25$ gives a discriminant of $601$, not a perfect square; $p=\sqrt5$ isn't even a valid rational coefficient itself.
Step 4: Conclude.
Only $p=-5$ works, matching option (2).
\[ \boxed{p=-5} \]