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

  1. #1
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    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
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  2. #2
    Super Member craig's Avatar
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    You were right with your roots,

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

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

    It has been a while since I have done inequalities so I might be wrong?
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  3. #3
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    Quote Originally Posted by adeshSB View Post
    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 $\displaystyle (x-3-\sqrt7)(x-3+\sqrt7)\le 0$
    In order that a product of two numbers be negative, the numbers must have opposite sign. So either $\displaystyle x-3-\sqrt{7}\le 0$ and $\displaystyle x-3+\sqrt{7}\ge 0$ or [tex]x-3-\sqrt{7}\ge 0[\math] and $\displaystyle x-3+\sqrt{7}\le 0$. In the first case $\displaystyle x\le 3+ \sqrt{7}$ and $\displaystyle x\ge 3- \sqrt{7}$: that is $\displaystyle 3-\sqrt{7}\le x\le 3+ \sqrt{7}$. In the second case, $\displaystyle x\ge 3+ \sqrt{7}$ and $\displaystyle x\le 3- \sqrt{7}$ which is impossible. All x such that $\displaystyle 3- \sqrt{7}\le x\le 3+ \sqrt{7}$.

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

    Finally, we can recognise that the graph of $\displaystyle y= x^2- 6x+ 2$ is a parabola opening upward. y will be less than 0 only for the portion of the parabola between the two roots $\displaystyle 3- \sqrt{7}$ and $\displaystyle 3+ \sqrt{7}$.
    Last edited by mr fantastic; Jan 8th 2009 at 06:25 PM. Reason: Fixed some latex ... itex --> math and [math] --> [/math]
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  4. #4
    Super Member craig's Avatar
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    Sorry you are quite right had a feeling there was something missing when I posted
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