+1859 Find all real numbers $x$ that satisfy the inequality $x^2+5x<6-x^2-3x$. Express your answer in interval notation.
+129771 Rearrange the inequality as
2x^2 + 8x - 6 < 0 divide through by 2
x^2 + 4x - 3 < 0
This does not factor
Let's turn it into an equality
x^3 + 4x - 3 = 0
The equation on the left is a parabola which turns upward....using the Quadratic Formula we can find the roots of this eqaution......the answer to the problem will lie between these roots (because the parbola will be below the x axis betwwen these values, e.g, it will be < 0 )
1st Root = ( -4 - sqrt [ 4^2 - (4 * 1 * -3) ] ) / 2 = [ -4 - sqrt ( 28) ] / 2 = [ -4 - 2sqrt (7) ] / 2 = -2 -sqrt (7)
The 2nd Root will be the conjugate of the first root = -2 + sqrt (7)
Solution x = ( -2 - sqrt (7) , -2 + sqrt (7) )
