How do you solve $0.23 x^{2} + 6.5 x + 4.3 < 0$ $0.23x^2+6.5x+4.3<0$ using a sign chart?

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Answer 1 · 2017-11-15T12:11:39

Solution : $- 27.58 < x < - 0.68 or x ∣ (- 27.58, - 0.68)$ $-27.58 < x < -0.68 or x| (-27.58,-0.68)$

$0.23 x^{2} + 6.5 x + 4.3 < 0$ $0.23x^2+6.5x+4.3<0$

Comparing with standard quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a = 0.23, b = 6.5, c = 4.3$ $a= 0.23 ,b=6.5 ,c=4.3$ Discriminant $D = b^{2} - 4 a c$ $D= b^2-4ac$ or

$D \approx 38.29$ $D ~~ 38.29$ If discriminant positive, we get two real solutions,

Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$ $x= (-b+-sqrtD)/(2a)$ or

$x= (-6.5+-sqrt38.29)/(2*0.23) :. x ~~ -27.58 , x ~~ -0.68$

$0.23x^2+6.5x+4.3<0$ or

$f(x)=0.23 (x +27.58)(x+0.68) <0$ .

Critical points are $x ~~ -27.58 , x ~~ -0.68$

Sign chart: When $x< -27.58$ sign of $f(x)$ is $(-) * (-) = (+) ; > 0$

When $-27.58 < x < -0.68$ sign of $f(x)$ is $(+) * (-) = (-) ; < 0$

When $x > -0.68$ sign of $f(x)$ is $(+) * (+) = (+) ; > 0$

Solution : $-27.58 < x < -0.68 or x| (-27.58,-0.68)$

graph{0.23x^2+6.5x+4.3 [-160, 160, -80, 80]}

[Ans]

How do you solve 0.23x^2+6.5x+4.3<00.23x2+6.5x+4.3<00.23x^2+6.5x+4.3<0 using a sign chart?