# can anyone help with my AS level revision i have a few practise quests. but no ans.

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• January 8th 2009, 11:52 AM
adeshSB
can anyone help with my AS level revision i have a few practise quests. but no ans.
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
• January 8th 2009, 01:37 PM
craig
You were right with your roots,

$x^2-6x+2$ roots are, , however the equation in factorised form is, $(x-3-\sqrt7)(x-3+\sqrt7)$.

I presume that the inequatities are therefore;
$x\leq3+\sqrt7$ and $x\leq3-\sqrt7$.

It has been a while since I have done inequalities so I might be wrong?
• January 8th 2009, 02:27 PM
HallsofIvy
Quote:

Originally Posted by adeshSB
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 $(x-3-\sqrt7)(x-3+\sqrt7)\le 0$
In order that a product of two numbers be negative, the numbers must have opposite sign. So either $x-3-\sqrt{7}\le 0$ and $x-3+\sqrt{7}\ge 0$ or [tex]x-3-\sqrt{7}\ge 0[\math] and $x-3+\sqrt{7}\le 0$. In the first case $x\le 3+ \sqrt{7}$ and $x\ge 3- \sqrt{7}$: that is $3-\sqrt{7}\le x\le 3+ \sqrt{7}$. In the second case, $x\ge 3+ \sqrt{7}$ and $x\le 3- \sqrt{7}$ which is impossible. All x such that $3- \sqrt{7}\le x\le 3+ \sqrt{7}$.

Another way to do this is to recognize that since $x^2- 6x+ 2$ 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 $3- \sqrt{7}$ and $3+\sqrt{7}$ break the real line into. $0< 3- \sqrt{7}$ and $0^2- 6(0)+ 2= 2> 0$ so no $x< 3- \sqrt{7}$ satisfies $x^2- 6x+ 2< 0$. $3-\sqrt{7}\le 3\le 3+\sqrt{7}$ and $(3)^2- 6(3)+ 2= -7< 0$ so every number between $3-\sqrt{7}$ and $3+ \sqrt{7}$ satisfies $x^2- 6x+ 2< 0$. Finally, $6> 3+ \sqrt{7}$ and $(6)^2- 6(6)+ 2= 2> 0$ so no number larger than $3+ \sqrt{7}$ satisfies the inequality.

Finally, we can recognise that the graph of $y= x^2- 6x+ 2$ is a parabola opening upward. y will be less than 0 only for the portion of the parabola between the two roots $3- \sqrt{7}$ and $3+ \sqrt{7}$.
• January 8th 2009, 02:32 PM
craig
Sorry you are quite right :) had a feeling there was something missing when I posted (Worried)
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