How do you solve $(2y^2+3y-20)/(y^3-y^2)>0$ ?

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How do you solve $(2y^2+3y-20)/(y^3-y^2)>0$ ?

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Answer 1 · 2017-01-11T10:29:47

The answer is $y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[$

Explanation:

Factorise the numerator and denominator

$2y^2+3y-20=(2y-5)(y+4)$

$y^3-y^2=y^2(y-1)$

So,

$(2y^2+3y-20)/(y^3-y^2)=((2y-5)(y+4))/(y^2(y-1))$

Let $y=((2y-5)(y+4))/(y^2(y-1))$

The domain of $y$ is $D_y=RR-{0,1}$

$y^2>0 AA y in D_y$

Let's do a sign chart

$color(white)(aaaa)$ $y$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaaaaaa)$ $0$ $color(white)(aaaa)$ $1$ $color(white)(aaaaa)$ $5/2$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $y+4$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)()$ $+$ $color(red)(∥)$ $color(white)(a)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $y-1$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)()$ $-$ $color(red)(∥)$ $color(white)(a)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $2y-5$ $color(white)(aaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)()$ $-$ $color(red)(∥)$ $color(white)(a)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(y)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)()$ $+$ $color(red)(∥)$ $color(white)(a)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(y)>0$ , when $y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[$

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How do you solve $(2y^2+3y-20)/(y^3-y^2)>0$ ?

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Explanation:

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How do you solve (2y^2+3y-20)/(y^3-y^2)>0(2y^2+3y-20)/(y^3-y^2)>0?

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Explanation:

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How do you solve $(2y^2+3y-20)/(y^3-y^2)>0$ ?