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

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Answer 1 · 2016-06-21T11:43:55

$x \in [- 4; \frac{3}{2}]$ $x in [-4;3/2]$

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

using the formula:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case you have:

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

so that

$x = \frac{- 5 \pm \sqrt{25 - 4 (2) (- 12)}}{2 (2)}$ $x=(-5+-sqrt(25-4(2)(-12)))/(2(2))$

$x = \frac{- 5 \pm \sqrt{25 + 96}}{4}$ $x=(-5+-sqrt(25+96))/4$

$x = \frac{- 5 \pm \sqrt{121}}{4}$ $x=(-5+-sqrt(121))/4$

$x = \frac{- 5 \pm 11}{4}$ $x=(-5+-11)/4$

$x = - \frac{16}{4} = - 4 and x = \frac{6}{4} = \frac{3}{2}$ $x=-16/4=-4 and x=6/4=3/2$

The trynomial is therefore negative in the interval

$[- 4; \frac{3}{2}]$ $[-4;3/2]$

How do you solve and write the following in interval notation: 2x2+5x−12≤02x^2+5x-12 ≤ 0?