How do you solve and write the following in interval notation: $2x^2+5x-12 ≤ 0$ ?

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Answer 1 · 2016-06-21T11:43:55

$x in [-4;3/2]$

You must find the zeroes of the trynomial
$2x^2+5x-12$

using the formula:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case you have:

$a=1;b=5;c=-12$

so that

$x=(-5+-sqrt(25-4(2)(-12)))/(2(2))$

$x=(-5+-sqrt(25+96))/4$

$x=(-5+-sqrt(121))/4$

$x=(-5+-11)/4$

$x=-16/4=-4 and x=6/4=3/2$

The trynomial is therefore negative in the interval

$[-4;3/2]$

How do you solve and write the following in interval notation: 2x^2+5x-12 ≤ 02x^2+5x-12 ≤ 0?