If $| x - 6 | > | x^{2} - 5 x + 9 |$ , then $1 < x < 3$

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Solution

(a) Here $x^{2} - 5 x + 9$ is always + ive
as $Δ = 25 - 36 = - i v e$ and hence its sign is same as that of 1st term. Hence the equation reduces to
$x^{2} - 5 x + 9 < | x - 6 |$
Now consider two cases x > 6 and x < 6.
1st case : $x^{2} - 5 x + 9 < x - 6, x > 6$
$x^{2} - 6 x + 15 < 0$
L.H.S. is always + ive as its $Δ$ is - ive and its sign is same as that of 1 st term. Hence this is ruled out as +ive cannot be < 0
2nd Case : $x^{2} - 5 x + 9 < - (x - 6), x < 6.$
or $x^{2} - 4 x + 3 < 0$ or $(x - 1) (x - 3) < 0$
$∴ 1 < x < 3.$

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| x^{2} + 5 x - 6 | = 6 - 5 x - x^{2}

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x^{2} + 5 x + 6

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A

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