Mathematics - Quadratic Equation Question with Solution | TestHub

Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given that one root is the negative of the other, we can write $\beta =-\alpha$. Therefore, the sum of the roots is $\alpha +\beta =\alpha +(-\alpha )=0$. From Vieta's formulas, the sum of the roots of $x^2-(a^2-4a+3)x-(a^2-5a+6)=0$ is $-(-(a^2-4a+3))=a^2-4a+3$. Equating the sum of roots to 0: $a^2-4a+3=0$. Factoring the quadratic: $(a-1)(a-3)=0$. This gives two possible integer values for $a$: $a=1$ or $a=3$. Now we must check if these values of $a$ lead to valid roots. The product of the roots is $\alpha \beta =\alpha (-\alpha )=-{\alpha }^2$. From Vieta's formulas, the product of the roots is $-(a^2-5a+6)$. So, $-{\alpha }^2=-(a^2-5a+6)$, which means ${\alpha }^2=a^2-5a+6$. For $a=1$: ${\alpha }^2=1^2-5(1)+6=1-5+6=2$. So $\alpha =\pm \sqrt{2}$. These are non-zero real roots. Thus, $a=1$ is a valid value. For $a=3$: ${\alpha }^2=3^2-5(3)+6=9-15+6=0$. So $\alpha =0$. In this case, both roots are 0. One root (0) is the negative of the other (0), so this condition is also satisfied. Thus, $a=3$ is a valid value. Both $a=1$ and $a=3$ satisfy the given condition. Therefore, the number of integer values of $a$ is 2.