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:

Roots are rational only if \(D \ge 0\) and \(D\) is a perfect square. If \(D\) is not a perfect square (like \(2\), \(3\), etc.), roots are irrational. If \(D<0\), roots are non-real.

Answer 1

The Correct Option is B

To determine the value of \( p \) for which the roots of the quadratic equation \(x^2 - px + 6 = 0\) are rational, we need to ensure that the discriminant of the equation is a perfect square.

The given quadratic equation is:

x^2 - px + 6 = 0

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), where for our equation, \(a = 1\), \(b = -p\), and \(c = 6\).

The discriminant \((D)\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

D = b^2 - 4ac

Substituting the values from the given quadratic:

D = (-p)^2 - 4 \times 1 \times 6

D = p^2 - 24

For the roots to be rational, \(D\) must be a perfect square. Let's assume \(D = k^2\) where \(k\) is an integer:

p^2 - 24 = k^2

Rearranging this gives:

p^2 - k^2 = 24

Recognize this as a difference of squares:

(p - k)(p + k) = 24

Now, we need integer factor pairs of 24:

• (1, 24)
• (2, 12)
• (3, 8)
• (4, 6)
• (-1, -24)
• (-2, -12)
• (-3, -8)
• (-4, -6)

Evaluate \((p - k, p + k) = (4, 6)\):

p - k = 4

p + k = 6

Adding the equations:

(p - k) + (p + k) = 4 + 6

2p = 10

p = 5

From the factor pair \((4, 6)\), we find that \() does not match the given options. Let's evaluate \((-3, -\):

p - k = -3

p + k = -8

Adding the equations:

(p - k) + (p + k) = -3 - 8

2p = -11

p = -5.5

Unfortunately, this also does not work. Finally, evaluate \((6, 4):

p - k = -4

p + k = -6

Adding the equations:

(p - k) + (p + k) = -4 - 6

2p = -10

p = -5

Therefore, the value of \( p \) for which the quadratic equation has rational roots is:

p = -5

Thus, the correct answer is -5, which matches one of the provided options.