You were right with your roots,
roots are, , however the equation in factorised form is, .
I presume that the inequatities are therefore;
and .
It has been a while since I have done inequalities so I might be wrong?
i attmepted this one and got as far as possible but not sure where to go next
find the solutions to x^2 -6x+2 leave answer as a surd i attempted it and ended up with 3+/- (sq. root of 7)
and then it asks to then get the exact solution to the inequality x^2-6x+2 </= 0
by substitution of (3+/- sq.root of 7) into x
i got to a point where -5</= 0
am i going right and does this mean X >/=5
Thanks for help
Unfortunately, craig was not completely right. The factorization craig gives makes your inequality
In order that a product of two numbers be negative, the numbers must have opposite sign. So either and or [tex]x-3-\sqrt{7}\ge 0[\math] and . In the first case and : that is . In the second case, and which is impossible. All x such that .
Another way to do this is to recognize that since is a continuous function so can change form negative to positive only where it is equal to 0. It is sufficient to check one value in each of the three intervals and break the real line into. and so no satisfies . and so every number between and satisfies . Finally, and so no number larger than satisfies the inequality.
Finally, we can recognise that the graph of is a parabola opening upward. y will be less than 0 only for the portion of the parabola between the two roots and .