Question

write a quadratic inequality for which the solution is -2 12 years ago

Answer 1

f (x) = ax^2 + bx + c < 0
(x + 2)(x -5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10 < 0.
The parabola is upward. Between the 2 real roots (-2) and (5), f(x) is negative as opposite in sign to a = 1. In this interval (-2, 5), the parabola is below the x-axis.

Answer 2

so wait the answer is x^2-3x-10<0?

Answer 3

to do that for say an equation, you'd use the values for "x" to get the roots, that is \(\bf x=-2\implies x+2=0\implies (x+2)=0
\\ \quad \\
x=5\implies x-5=0\implies (x-5)=0
\\ \quad \\
(x+2)(x-5)=0\)

Answer 4

in this case is an inequality, so you could write it as \(\bf (x+2)(x-5)<0\qquad (x+2)(x-5)>0\)

Answer 5

which is euqal to x^2-3x-10>0 right?

Answer 6

either one

so the quadratic equation will be the multiplication of both of the roots

yes, which is equals to \(\bf (x+2)(x-5)<0\implies x^2-3x-10<0
\\ \quad \\
(x+2)(x-5)>0\implies x^2-3x-10>0\)

so use the form that gives you the values back as you need them

Answer 7

for this case, it'd be the 1st one, because \(\bf (x+2)(x-5)<0\implies x^2-3x-10<0
\\ \quad \\
x^2-3x-10<0\implies (x+2)(x-5)<0\to
\begin{cases}
x+2<0\implies x<-2
\\ \quad \\
\bf x-5<0\implies x<5
\end{cases}\)

Answer 8

ok i think i get it :D thanks, do u know alot about like radians and stuff?

Answer 9

just post anew, thus more exposure, and we can revise each other

Answer 10

ok thanks :D

Answer 11

hmm just a quick revision \(\bf x+2>0\implies x>-2
\\ \quad \\
x<5\implies x-5<0\implies -x-5>0
\\ \quad \\
(x+2)(-x-5)>0\)

Answer 12

rather \(\bf x+2>0\implies x>-2
\\ \quad \\
x<5\implies x-5<0\implies -x+5>0
\\ \quad \\
(x+2)(-x+5)>0\)