How do you solve $2x^2+4x<=3$ using a sign chart?

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Answer 1 · 2017-01-04T07:07:11

The answer is $x in [-2.58, 0.58]$

Let`s rewrite the equation

$2x^2+4x-3<=0$

We need the roots of the equation

$2x^2+4x-3=0$

We calculate the discriminant

$Delta=b^2-4ac=16+4*2*3=40$

As $Delta>0$ , there are 2 real roots

Therefore,

$x_1=(-4+sqrt40)/4=0.58$

$x_2=(-4-sqrt40)/4=-2.58$

Let $f(x)=(x+2.58)(x-0.58)<=0$

Now, we can do our sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2.58$ $color(white)(aaaaa)$ $0.58$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+2.58$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaaa)$ $+$

$color(white)(aaaa)$ $x-0.58$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaaa)$ $+$

Therefore,

$f(x)<=0$ , when $x in [-2.58, 0.58]$

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