"The number of elements in the set {x $\in$ R : (|x| $-$ 3) |x + 4| = 6} is equal to :"

Question

Accepted Answer

"Case 1 :

x $\le$ $-$4

($-$x $-$ 3)($-$x $-$ 4) = 6

$\Rightarrow$ (x + 3)(x + 4) = 6

$\Rightarrow$ x² + 7x + 6 = 0

$\Rightarrow$ x = $-$1 or $-$6

but x $\le$ $-$4

x = $-$6

Case 2 :

x $\in$ ($-$4, 0)

($-$x $-$ 3)(x + 4) = 6

$\Rightarrow$ $-$x² $-$ 7x $-$ 12 $-$ 6 = 0

$\Rightarrow$ x² + 7x + 18 = 0

D < 0 No solution

Case 3 :

x $\ge$ 0

(x $-$ 3)(x + 4) = 6

$\Rightarrow$ x² + x $-$ 12 $-$ 6 = 0

$\Rightarrow$ x² + x $-$ 18 = 0

x = ${{ - 1 \pm \sqrt {1 + 72} } \over 2}$

$\therefore$ x = ${{\sqrt {73} - 1} \over 2}$ only"

The number of elements in the set {x $\in$ R : (|x| $-$ 3) |x + 4| = 6} is equal to :

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The number of elements in the set {x $\in$ R : (|x| $-$ 3) |x + 4| = 6} is equal to :

Solution

About this question

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