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 answerB$4$
- C$2$
- DNone
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: $4$
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'.