Question

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

Answer 1

Answer:

b. $x\approx -2.426$

Step-by-step explanation:

The given equation is;

$x^3+x^2+10x+15=0$

We solve by the x-intercept method. We need to graph the corresponding function using a graphing tool.

The corresponding function is

$f(x)=x^3+x^2+10x+15$

The solution to $x^3+x^2+10x+15=0$ 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:

$x\approx -2.426$

Answer 2

Answer:

b. x ≈ -2.426

Step-by-step explanation:

Given that we have possible roots we can replace these values into the equation and check if it is satisfied.

option a: 2.426^3+4*2.426^2+10*2.426+15 = 77.08 ≠ 0

option b: (-2.426)^3+4*(-2.426)^2+10*(-2.426)+15 ≈ 0

option c: 5.128^3+4*5.128^2+10*5.128+15 = 306.31 ≠ 0

option d: (-5.128)^3+4*(-5.128)^2+10*(-5.128)+15 = -65.94 ≠ 0