Let f(x)=x2+bx+9 and g(x)=x2+dx+e. If f(x)=0 has roots r and s, and g(x)=0 has roots −r and −s, compute the two roots of f(x)+g(x)=0.
Solution — click to reveal
We have that f(x)=(x−r)(x−s) and g(x)=(x+r)(x+s), so f(x)+g(x)=(x−r)(x−s)+(x+r)(x+s)=x2−(r+s)x+rs+x2+(r+s)x+rs=2x2+2rs=2(x2+rs).By Vieta's formulas, rs=9, so f(x)+g(x)=2(x2+9). Ths roots of x2+9=0 are 3i,−3i.