Discriminant vs Graphs

Algebra Level 2

f(x)=(xa)(xe)+ (xb)(xf)+ (xc)(xg)+ (xd)(xh)\begin{aligned} f(x) &= (x-a)(x-e)\\ &+ \ (x-b)(x-f)\\ &+ \ (x-c)(x-g)\\ &+ \ (x-d)(x-h) \end{aligned} For real numbers a<b<c<d<e<f<g<ha<b<c<d<e<f<g<h, we define the function as above. Then what can we say about the roots of f(x)=0f(x) = 0 ?

×

Problem Loading...

Note Loading...

Set Loading...