Question

If one of the roots of x(x + 2) = 4 - (1-ax²) tends to infinity, then a will tend to -

• A. 0
• B.
  • 1
• C. 1
• D. 2

Answer 1

Answer: (C)

1

Answer 2

The given equation is

x(x+2)=4-(1-ax^2)

Expanding the left side:

x^2+2x = 4 - 1 + ax^2 = 3 + ax^2

Rearrange the equation:

x^2+2x-ax^2-3=0 \quad \Longrightarrow \quad (1-a)x^2+2x-3=0

Let the quadratic be written as:

(1-a)x^2 + 2x - 3 = 0

For one of the roots to tend to infinity, the denominator in the quadratic formula must tend to zero, i.e., the coefficient of $x^2$ should tend to zero without canceling the corresponding term in the numerator.

This gives:

1-a \to 0 \quad \Longrightarrow \quad a\to 1

Verification:

Using the quadratic formula:

x=\frac{-2 \pm \sqrt{4+12(1-a)}}{2(1-a)}

As $a\to1$, i.e., $1-a\to0$:

• The “$+$” branch becomes $\frac{-2+2}{2(1-a)} = \frac{0}{2(1-a)} = 0$ (finite), and

• The “$-$” branch becomes $\frac{-2-2}{2(1-a)} = \frac{-4}{2(1-a)} = \frac{-2}{1-a}$ which tends to $-\infty$.

Thus, one root tends to infinity when $a$ tends to 1.