Mathematics - Quadratic Equation Question with Solution | TestHub

For the roots of the equation to be equal, its discriminant must be zero. $D=[-2b(a+c){]}^2-4(a^2+b^2)(b^2+c^2)=0$. This simplifies to $4b^2(a+c{)}^2-4(a^2{b}^2+a^2{c}^2+b^4+b^2{c}^2)=0$. Dividing by 4 and expanding: $b^2(a^2+2ac+c^2)-(a^2{b}^2+a^2{c}^2+b^4+b^2{c}^2)=0$. This further simplifies to $a^2{b}^2+2a{b}^{2}c+b^2{c}^2-a^2{b}^2-a^2{c}^2-b^4-b^2{c}^2=0$. Combining terms, we get $2a{b}^{2}c-a^2{c}^2-b^4=0$. Rearranging, $b^4-2a{b}^{2}c+a^2{c}^2=0$, which is $(b^2-ac{)}^2=0$. Thus, $b^2=ac$. This is the condition for $a,b,c$ to be in Geometric Progression (GP).