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Thread: equation inequalities

  1. #1
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    equation inequalities

    Find the solutions sets for the following inequalities.
    (a) |x − 2| < 4
    (b) $\displaystyle x^2 − 1 ≥ 3$
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  2. #2
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    Hello quah13579
    Quote Originally Posted by quah13579 View Post
    Find the solutions sets for the following inequalities.
    (a) |x − 2| < 4
    (b) $\displaystyle x^2 − 1 ≥ 3$
    (a) $\displaystyle |x-2| < 4$ means that the numerical value of $\displaystyle (x-2)$ is less than $\displaystyle 4$. So $\displaystyle (x-2)$ lies between $\displaystyle -4$ and $\displaystyle 4$.

    $\displaystyle \Rightarrow -4 < x-2 < 4$

    Take the LH inequality: $\displaystyle -4< x-2$

    $\displaystyle \Rightarrow -2<x$

    ...and the RH inequality: $\displaystyle x-2<4$

    $\displaystyle \Rightarrow x < 6$

    So, combining these two answers into one: $\displaystyle -2<x<6$

    (b) Although you had a LaTeX error in your original posting, I assume it's $\displaystyle x^2 - 1 \ge 3$

    So $\displaystyle x^2 \ge 4$

    When we take square roots, be careful with the negative values. We need to use an absolute value sign |...|, like this:

    $\displaystyle |x| \ge 2$

    $\displaystyle \Rightarrow x \le -2$ or $\displaystyle x \ge 2$

    Grandad
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  3. #3
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    [quote=quah13579;348363]Find the solutions sets for the following inequalities.
    (a) |x − 2| < 4
    (b) x^2 − 1 ≥ 3
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  4. #4
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    Find the solutions sets for the following inequalities.
    (a) |x − 2| < 4
    (b) x^2 − 1 ≥ 3
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by quah13579 View Post
    Find the solutions sets for the following inequalities.
    (a) |x − 2| < 4
    (b) $\displaystyle x^2 − 1 ≥ 3$
    for the absolute value

    $\displaystyle a)\mid x-2 \mid < 4$ in general

    $\displaystyle \mid x-a \mid < b \Rightarrow -b<x-a <b$
    and

    $\displaystyle \mid x-a \mid > b \Rightarrow -b>x $ or $\displaystyle b<x$
    so

    $\displaystyle \mid x-2 \mid < 4 \Rightarrow -4<x-2<4 \Rightarrow -4+2 < x < 4+2 \Rightarrow -2<x<6$

    second one

    $\displaystyle b)x^2-1 \geq 3 $

    $\displaystyle x^2-1-3 \geq 0 $

    $\displaystyle x^2-4 \geq 0 $

    $\displaystyle (x-2)(x+2) \geq 0 $ we want the values of x which make the value of $\displaystyle x^2-4$ positive or zero so we need to study the sign of it
    the root of it is -2 ,2 plot them in the real line like this
    equation inequalities-1.jpg

    then take numbers from the three intervals (-infinity , -2 ) and from (-2,2) and from (2,infinity) then sub them in the formula if it was positive write positive sign if negative write negative on the real line like this
    equation inequalities-2.jpg

    the solution is $\displaystyle x\in (\infty,-2] \cup [2,\infty)$
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