Discriminant vs Graphs

Algebra Level 2

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

purely imaginary do not exist imaginary and distinct real and distinct real and equal

1 solution

Julian Poon
Jun 10, 2015

$f(x)$ is a parabola, which means it has a line of symmetry. So, it either has $2$ real roots, repeated roots, or $2$ imaginary roots (which means $f(x)$ does not touch the x-axis). We can see that $\text{When } x<a, f(x)<0$ $\text{When } x>h, f(x)>0$ This shows that $f(x)$ crossed at x-axis, and will never be tangent nor not cross at all with the x-axis. Therefore, we can conclude that it's roots are $\boxed{\text{ real and distinct}}$

Your first staement is false. When $x<a$ , we instead have $f(x)>0$ (product of two negatives in each quadratic); you should mean $d<x<e$ instead for $f(x)<0$

Jared Low - 5 years, 6 months ago

Discriminant vs Graphs

1 solution

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