Question

If the quadratic equation $x^2+ax+b=0$ and $x^2+bx+a=0\left(a\ne b\right)$ have a common root, then find the numerical value of $a+b$

Answer 1

If $a,x^2+b,x+c_1=0$

and $a_2{x}^2+b^{2}x+c_2=0$

have a common root, then conditions for common root is

${\left|\begin{array}{cc}{1}_1& c_1\\ a_2& c_2\end{array}\right|}^2=\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|\left|\begin{array}{cc}{b}_1& c_1\\ b_2& c_2\end{array}\right|$

In case ,equation are

$x^2+ax+b=0$

and $x^2+bx+a=0$

$\therefore$ for common root

${\left|\begin{array}{cc}1& b\\ 1& a\end{array}\right|}^2=\left|\begin{array}{cc}1& a\\ 1& b\end{array}\right|\left|\begin{array}{cc}a& b\\ b& a\end{array}\right|$

$(a-b{)}^2=(b-a\left)\right(a^2-b^2)$

$(a-b{)}^2=-(a-b\left)\right(a-b\left)\right(a+b)$

$(a-b{)}^2=-(a-b{)}^2(a+b)$

or $(a-b{)}^2+(a-b{)}^2(a+b)=0$

or $(a-b{)}^2[1+a+3]=0$

$(a-b{)}^2=0$

$a=b$

but $\therefore a\ne b$ as equation become same

$1+a+b+=0$

or $a+b=-1$