Question

hello friends please help me ​

Answer 1

Step-by-step explanation:

Step-by-step explanation:

$\sf \frac { 6{x}^{2} -5x - 3 }{ {x}^{2} - 2x + 6 } < 4 \\ \\ \sf \frac { 6{x}^{2} -5x - 3 }{ {x}^{2} - 2x + 6 } - 4 < 0 \\ \\ \sf \frac { 6{x}^{2} -5x - 3 - 4({x}^{2} -2x + 6) }{ {x}^{2} - 2x + 6 } < 0 \\ \\\sf \frac { 6{x}^{2} -5x - 3 - 4{x}^{2} +8x -24 }{ {x}^{2} - 2x + 6 } < 0\\ \\ \sf \frac { 2{x}^{2} +3x - 27}{ {x}^{2} - 2x + 6 } < 0 \\ \\ \sf Let \:us\:check\:the\: \Delta \: for \:the \:denominator \\ \\ \sf \Delta = {b}^{2} - 4ac \\ \\ \sf = {2}^{2} - 4 × 6 \\ \\ \sf = - 22 \\ \\ \sf <0 \\ \\ \sf Hence\:the\: denominator\:is\:always\:positive \\ \\ \sf \therefore 2{x}^{2} + 3x - 27 < 0 \\ \\ \sf \therefore 2{x}^{2} - 6x + 9x - 27 < 0 \\ \\ \sf \therefore 2x ( x - 3 ) + 9 ( x - 3 ) < 0 \\ \\ \sf \therefore ( x - 3 )( 2x + 9 ) < 0 \\ \\ \sf Now\:using\:wavy\:curve\:method\: \\ \\ \sf x \belongs [ - \frac {9}{2} , 3 ] \\ \\ \sf Hence\: option\:A \:is \:correct$

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Answer 2

Answer:

Step-by-step explanation:

\begin{gathered}\sf \frac { 6{x}^{2} -5x - 3 }{ {x}^{2} - 2x + 6 } < 4 \\ \\ \sf \frac { 6{x}^{2} -5x - 3 }{ {x}^{2} - 2x + 6 } - 4 < 0 \\ \\ \sf \frac { 6{x}^{2} -5x - 3 - 4({x}^{2} -2x + 6) }{ {x}^{2} - 2x + 6 } < 0 \\ \\\sf \frac { 6{x}^{2} -5x - 3 - 4{x}^{2} +8x -24 }{ {x}^{2} - 2x + 6 } < 0\\ \\ \sf \frac { 2{x}^{2} +3x - 27}{ {x}^{2} - 2x + 6 } < 0 \\ \\ \sf Let \:us\:check\:the\: \Delta \: for \:the \:denominator \\ \\ \sf \Delta = {b}^{2} - 4ac \\ \\ \sf = {2}^{2} - 4 × 6 \\ \\ \sf = - 22 \\ \\ \sf < 0 \\ \\ \sf Hence\:the\: denominator\:is\:always\:positive \\ \\ \sf \therefore 2{x}^{2} + 3x - 27 < 0 \\ \\ \sf \therefore 2{x}^{2} - 6x + 9x - 27 < 0 \\ \\ \sf \therefore 2x ( x - 3 ) + 9 ( x - 3 ) < 0 \\ \\ \sf \therefore ( x - 3 )( 2x + 9 ) < 0 \\ \\ \sf Now\:using\:wavy\:curve\:method\: \\ \\ \sf x \belongs [ - \frac {9}{2} , 3 ] \\ \\ \sf Hence\: option\:A \:is \:correct \end{gathered}

x

2

−2x+6

6x

2

−5x−3

<4

x

2

−2x+6

6x

2

−5x−3

−4<0

x

2

−2x+6

6x

2

−5x−3−4(x

2

−2x+6)

<0

x

2

−2x+6

6x

2

−5x−3−4x

2

+8x−24

<0

x

2

−2x+6

2x

2

+3x−27

<0

LetuschecktheΔforthedenominator

Δ=b

2

−4ac

=2

2

−4×6

=−22

<0

Hencethedenominatorisalwayspositive

∴2x

2

+3x−27<0

∴2x

2

−6x+9x−27<0

∴2x(x−3)+9(x−3)<0

∴(x−3)(2x+9)<0

Nowusingwavycurvemethod

x\belongs[−

2

9

,3]

HenceoptionAiscorrect

Step-by-step explanation:

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