Question

Sum of the real values of ${}^{\prime }{a}^{\prime }$ for which the equation $(a^2-3a+2)x^2+(a^2-4a+3)x+(a^2-6a+5)=0$ has three distinct roots

• A. $11$
• B. $1$
• C. $10$
• D. None

Answer 1

The correct option is B

$1$

The quadratic equation has three distinct roots implies it is an identity.

Then $a^2-3a+2=0$

$\Rightarrow a^2-2a-a+2=0$

$\Rightarrow a(a-2)-1(a-2)=0$

$\Rightarrow (a-1)(a-2)=0$

$\Rightarrow a=1,2$


$a^2-4a+3=0$

$\Rightarrow a^2-3a-a+3=0$

$\Rightarrow a(a-3)-1(a-3)=0$

$\Rightarrow (a-1)(a-3)=0$

$\Rightarrow a=1,3$

$a^2-6a+5=0$

$\Rightarrow a^2-5a-a+5=0$

$\Rightarrow a(a-5)-1(a-5)=0$

$\Rightarrow (a-1)(a-5)=0$

$\Rightarrow a=1,5$

$\therefore a=1$.