# Thread: Finding the real solutions

1. ## Finding the real solutions

Hi

Can someone help me how to solve the following:

1) Find the real solutions of the cubic equation:$\displaystyle 9x^3 + 8x +6$

This is what i have done:

$\displaystyle let x = x$

$\displaystyle 9x^2 +8+6 = 0$

$\displaystyle 9x^2 = -14$

$\displaystyle x^2=\frac{-14}{9}$

$\displaystyle x=1.2472$

2) Find the inequality corresponding to the graph:

P.S

2. Originally Posted by Paymemoney
Hi

Can someone help me how to solve the following:

1) Find the real solutions of the cubic equation:$\displaystyle 9x^3 + 8x +6$

This is what i have done:

$\displaystyle let x = x$ , what do u mean ?

$\displaystyle 9x^2 +8+6 = 0$ , how did the x^3 become x^2

$\displaystyle 9x^2 = -14$

$\displaystyle x^2=\frac{-14}{9}$ , you can't have negative values in the square root

$\displaystyle x=1.2472$ , this is wrong

hi

this can't be solved algebraically (Perhaps it can , but i don see how). The fastest way to solve it is by using a calculator OR you can also approximate the roots using the newton-raphson method .

hi

this can't be solved algebraically (Perhaps it can , but i don see how). The fastest way to solve it is by using a calculator OR you can also approximate the roots using the newton-raphson method .

ok thanks,

also the second question forgot to add that there were 4 answers to choose from:

$\displaystyle y>0.272x+0.428-\frac{2}{x}$

$\displaystyle y>-0.419x-0.360$

$\displaystyle y>-1.291x-1.45$

$\displaystyle y>-0.521x+0.603-\frac{1}{x}$

$\displaystyle y>1.97x-0.142-\frac{1}{x^2}$

4. It should be obvious that the lower boundary of that region is a straight line and so its equation is linear. That excludes all except the second and third. In the picture it looks like when x= 0, y is between -.25 and -.50. Which of the given options is that?