Q161 / 270 · NDA 2026 (Phase 1) · Mathematics

If the highest degree coefficient is equal to $1$, then what is the total number of quadratic equations which are unchanged on squaring their roots ?

  • A
    $6$
  • Correct answer
    B
    $4$
  • C
    $2$
  • D
    None

Solution

Let the quadratic equation be $x^2 + px + q = 0$.

Let its roots be $\alpha$ and $\beta$.

Sum of roots: $\alpha + \beta = -p$

Product of roots: $\alpha \beta = q$

The roots of the new equation are $\alpha^2$ and $\beta^2$.

Sum of new roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$

Product of new roots: $\alpha^2 \beta^2 = q^2$

Since the equation remains unchanged, the sum and product of the roots must be the same as the original equation.

Therefore, $p^2 - 2q = -p$ and $q^2 = q$.

From $q^2 = q$, we get $q = 0$ or $q = 1$.

Case 1: $q = 0$

Substituting $q = 0$ in $p^2 - 2q = -p$, we get $p^2 = -p \Rightarrow p^2 + p = 0 \Rightarrow p(p + 1) = 0$.

This gives $p = 0$ or $p = -1$.

The corresponding equations are $x^2 = 0$ and $x^2 - x = 0$.

Case 2: $q = 1$

Substituting $q = 1$ in $p^2 - 2q = -p$, we get $p^2 - 2 = -p \Rightarrow p^2 + p - 2 = 0 \Rightarrow (p + 2)(p - 1) = 0$.

This gives $p = -2$ or $p = 1$.

The corresponding equations are $x^2 - 2x + 1 = 0$ and $x^2 + x + 1 = 0$.

Thus, there are $4$ such quadratic equations.

Answer: $4$

Hence, the correct answer is option: 'B'.

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