Question

The number of integral values of m for which the equation $(1 + m^2)x^2 - 2(1 + 3m)x + (1 + 8m) = 0$ has no real root is :

• A. infinitely many
• B. 2
• C. 3
• D. 1

Answer 1

The Correct Option is A

We are given the quadratic equation \((1 + m^2)x^2 - 2(1 + 3m)x + (1 + 8m) = 0\). To determine the number of integral values of \(m\) for which this equation has no real roots, we need to analyze the discriminant of the quadratic equation. 

The quadractic equation \(ax^2 + bx + c = 0\) has real roots if and only if its discriminant \(b^2 - 4ac\) is non-negative. Therefore, for the equation to have no real roots, the discriminant should be less than zero. Let's calculate the discriminant:

• The coefficient \(a = 1 + m^2\)
• The coefficient \(b = -2(1 + 3m) = -2 - 6m\)
• The constant term \(c = 1 + 8m\)

The discriminant \(\Delta\) is given by:

\(\Delta = b^2 - 4ac = (-2 - 6m)^2 - 4(1 + m^2)(1 + 8m)\)

Expanding and simplifying:

• \((-2 - 6m)^2 = 4 + 24m + 36m^2\)
• \(4(1 + m^2)(1 + 8m) = 4(1 + 8m + m^2 + 8m^3) = 4 + 32m + 4m^2 + 32m^3\)

Now, substitute these into the discriminant expression:

\(\Delta = 4 + 24m + 36m^2 - (4 + 32m + 4m^2 + 32m^3)\)

Simplify to find:

\(\Delta = 36m^2 - 4m^2 + 24m - 32m - 32m^3 = -32m^3 + 32m + 32m^2\)

\(\Delta = -32m^3 + 32m^2 - 8m\)

We need \(\Delta < 0\) for no real roots:

\(-32m^3 + 32m^2 - 8m < 0\)

Factor out the common term:

\(-8m(4m^2 - 4m + 1) < 0\)

Now, solve the inequality \(4m^2 - 4m + 1 > 0\). Notice that this is a perfect square. Hence:

\((2m - 1)^2 \geq 0\) for all real \(m\).

The perfect square inequality never becomes negative, thus it affects the signs such that the inequality \(-8m((2m - 1)^2) < 0\) is true for \(m > 0\) as the leading term is negative, making the whole expression less than zero.

Thus, all positive integer \(m\) values satisfy this inequality, which implies infinitely many values.

Therefore, the correct answer is: infinitely many.