Question

the number of integral values of x satisfying the equation x^2-|x| -6 less than equal to zero is​

Answer 1

Given info : equation is x² - |x| - 6 ≤ 0

To find : the number of integral solutions of x.

Solution : x² - |x| - 6 ≤ 0

⇒|x|² - |x| - 6 ≤ 0 [ we can use x² = |x|² ]

⇒|x|² - 3|x| + 2|x| - 6 ≤ 0

⇒|x|(|x| - 3) + 2(|x| - 3) ≤ 0

⇒(|x| + 2)(|x| - 3) ≤ 0

⇒-2 ≤ |x| ≤ 3

Case 1 : |x|≥ -2

It is true for all real value of x.

Case 2 : |x| ≤ 3

-3 ≤ x ≤ 3

Then, x belongs to [-3, 3]

x = -3, -2, -1, 0, 1, 2, 3

Therefore are seven integral solutions of x satisfying the equation x² - |x| - 6 ≤ 0

Answer 2

SOLUTION :

TO DETERMINE

The number of integral values of x satisfying the equation

$\sf{ {x}^{2} - |x| - 6 \leqslant 0\: }$

EVALUATION

The given equation is

$\sf{ {x}^{2} - |x| - 6 \leqslant 0\: } \: \: .....(1)$

$\sf{ CASE \: : 1 \: \: ( \: Let \: x > 0 \: ) }$

Then from (1) we get

$\sf{ {x}^{2} - x - 6 \leqslant 0\: }$

$\implies \sf{ {x}^{2} - 3 x + 2x - 6 \leqslant 0\: }$

$\implies \sf{ (x - 3) (x + 2)\leqslant 0\: }$

$\implies \sf{ - 2 \leqslant x \leqslant 3\: }$

$\sf{Since \: we \: assumed \: that \: \: x > 0 }$

So we have

$\sf{ 0 < x \leqslant 3\: } \: \: ....(2)$

$\sf{ CASE \: : 2 \: \: ( \: Let \: x \leqslant 0 \: ) }$

The from (1) we get

$\sf{ {x}^{2} + x - 6 \leqslant 0\: }$

$\implies \sf{ {x}^{2} + 3x - 2x - 6 \leqslant 0\: }$

$\implies \sf{ (x + 3)(x - 2) \leqslant 0\: }$

$\implies \sf{ - 3 \leqslant x \leqslant 2\: }$

$\sf{Since \: we \: assumed \: that \: \: x \leqslant 0 }$

So we have

$\sf{ - 3 \leqslant x \leqslant 0\: } \: \: ....(3)$

(2) and (3) together gives

$\sf{ - 3 \leqslant x \leqslant 3}$

So the solution set is

$\sf{ \{ \: \: x \in \mathbb{Z} : \: - 3 \leqslant x \leqslant 3 \: \} }$

$\sf{And \: the \: solutions \: are \: - 3, - 2, - 1, 0, 1,2,3 \: }$

Hence there are 7 integral values of x satisfying the equation

$\sf{ {x}^{2} - |x| - 6 \leqslant 0\: }$

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