Question

If the equations $a{x}^2+bx+c=0$ and $c{x}^2+bx+a=0$ have one root in common, then $(a+b+c)\times (a-b+c)$ equal to:

Answer 1

$forstandardequationsoftheform{a}_1{x}^2+b_{1}x+c_1=0and{a}_2{x}^2+b_{2}x+c_2=0conditionforcommonrootis(c_1{a}_2-c_2{a}_1{)}^2=(b_1{c}_2-b_2{c}_1\left)\right(a_1{b}_2-a_2{b}_1\left)\right(c\cdot c-a\cdot a{)}^2=(ba-bc)(ab-cb)(c^2-a^2{)}^2=-b(c-a\left)\right(-b\left)\right(c-a\left)\right(c^2-a^2{)}^2-b^2(c-a{)}^2=0(c-a{)}^2\left[\right(c+a{)}^2-b^2]=0asc\ne a\therefore (c+a-b\left)\right(c+a+b)=0$