Question

Solve: 144y^4 - 337y^2 + 144= 0​

Answer 1

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Answer 2

$144y^4-337y^2+144=0\quad :\quad y=\frac{4}{3},\:y=-\frac{4}{3},\:y=\frac{3}{4},\:y=-\frac{3}{4}$

Solution:

Given that,

$144y^4-337y^2+144 = 0\\\\\mathrm{Let\:}u=y^2\\\\144u^2-337u+144 =0$

$\mathrm{Factor}\:144u^2-337u+144$

$\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(144u^2-81u\right)+\left(-256u+144\right) = 0\\\\\mathrm{Factor\:out\:}9u\mathrm{\:from\:}144u^2-81u\mathrm{:\quad }9u\left(16u-9\right)$

$\mathrm{Factor\:out\:}-16\mathrm{\:from\:}-256u+144\mathrm{:\quad }-16\left(16u-9\right)\\\\9u\left(16u-9\right)-16\left(16u-9\right) = 0\\\\\mathrm{Factor\:out\:common\:term\:}16u-9\\\\\left(16u-9\right)\left(9u-16\right) = 0\\\\\mathrm{Substitute\:back}\:u=y^2\\\\\left(16y^2-9\right)\left(9y^2-16\right) = 0$

$\mathrm{Factor}\:16y^2-9:\quad \left(4y+3\right)\left(4y-3\right)\\\\Therefore\\\\\left(4y+3\right)\left(4y-3\right)\left(9y^2-16\right) = 0\\\\\mathrm{Factor}\:9y^2-16:\quad \left(3y+4\right)\left(3y-4\right)\\\\Therefore\\\\\left(4y+3\right)\left(4y-3\right)\left(3y+4\right)\left(3y-4\right) = 0\\\\Therefore,\\\\y = \pm \frac{3}{4} \\\\y = \pm \frac{4}{3}$

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