How to start?

Question

<Inequality>
How to start?

callisto · Answer

attachment

amistre64 · Answer

solve the quadractic

$$kx^2-ax+b~<~0$$

amistre64 · Answer

spose to be a -b but yeah

callisto · Answer

$$kx^2-ax-b<0$$Since they have intercepts at x=c and x=d, so, would the solution be
c<x<d but that doesn't seem to be right..

anonymous · Answer

isnt that just looking at figure, its C @.@ am i missing something?

amistre64 · Answer

this equals zero when:
$$x=\frac{a}{2k}\pm\frac{\sqrt{a^2+4bk}}{2k}$$

callisto · Answer

No o_o
c<x  x>d o_o

amistre64 · Answer

the figure itself is a generalization so we cant rely on what we see

amistre64 · Answer

$$c=\frac{a}{2k}-\frac{\sqrt{a^2+4bk}}{2k}$$
$$d=\frac{a}{2k}+\frac{\sqrt{a^2+4bk}}{2k}$$

amistre64 · Answer

if we are allowed to use the figure as a true rendering, then yes  less then c and greater than d would be accurate

callisto · Answer

Summarize:
$$kx^2 < ax+b$$$$kx^2 - ax-b <0$$Since they have intercepts at x=c and x=d, so
x<c or x>d
That would give us the answer C.

How can I get the answer by looking at the graph?

amistre64 · Answer

find all the places where kx^2 is lower, under, beneath, the line of ax+b

amistre64 · Answer

the range of the values:  y values

for which kx^2  is less than ax+b|dw:1371736732039:dw|