Question

Number of values $p$ for which the equation $(p^2-3p+2)x^2-(p^2-5p+4)x+p-p^2=0$ possess more than two roots, is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

A polynomial equation of degree $n$ has at most $n$ roots. For a quadratic equation (degree 2) to have more than two roots, it must be an identity, meaning it holds true for all values of $x$. This occurs when all the coefficients of the polynomial are zero.

Let the given equation be $Ax^2 + Bx + C = 0$, where: $A = p^2-3p+2$ $B = -(p^2-5p+4)$ $C = p-p^2$

We need to find the values of $p$ for which $A=0$, $B=0$, and $C=0$ simultaneously.

Factorize each coefficient:

1. $A = p^2-3p+2 = (p-1)(p-2)$ For $A=0$, $p=1$ or $p=2$.

2. $B = -(p^2-5p+4) = -(p-1)(p-4)$ For $B=0$, $p=1$ or $p=4$.

3. $C = p-p^2 = p(1-p) = -p(p-1)$ For $C=0$, $p=0$ or $p=1$.

For all three coefficients to be zero, $p$ must be a common value in the solutions of $A=0$, $B=0$, and $C=0$. The common value is $p=1$.

When $p=1$: $A = (1-1)(1-2) = 0 \times (-1) = 0$ $B = -(1-1)(1-4) = -0 \times (-3) = 0$ $C = -1(1-1) = -1 \times 0 = 0$

Thus, for $p=1$, the equation becomes $0x^2 + 0x + 0 = 0$, which is $0=0$. This equation is true for all real numbers $x$, meaning it has infinitely many roots. The question asks for the number of values of $p$ for which this condition holds. We found only one such value, $p=1$.