MATH SOLVE

3 months ago

Q:
# Find all real solutions of the equation, approximating when necessary.x^3+4x^2=10x+15=0

Accepted Solution

A:

Answer:b. [tex]x\approx -2.426[/tex]Step-by-step explanation:The given equation is; [tex]x^3+x^2+10x+15=0[/tex]We solve by the x-intercept method. We need to graph the corresponding function using a graphing tool.The corresponding function is [tex]f(x)=x^3+x^2+10x+15[/tex]The solution to [tex]x^3+x^2+10x+15=0[/tex] is where the graph touches the x-axis.We can see from the graph that; the x-intercept is;(-2.426,0)Therefore the real solution is:[tex]x\approx -2.426[/tex]