Question

Solve the cube root equation x3+14x2+49x+20−−−−−−−−−−−−−−−−−√3=x+5
𝑥
3
+
14
𝑥
2
+
49
𝑥
+
20
3
=
𝑥
+
5
.(1 point)
Responses

x=−21
𝑥
=
−
21
and x=−5
𝑥
=
−
5
x equals negative 21 and x equals negative 5

x=5
𝑥
=
5
and x=21
𝑥
=
21
x equals 5 and x equals 21

x=−35
𝑥
=
−
35
and x=−3
𝑥
=
−
3
x equals negative 35 and x equals negative 3

x=3
𝑥
=
3
and x=35

Answer 1

To solve the equation

\[
\sqrt[3]{x^3 + 14x^2 + 49x + 20} = x + 5,
\]

we can start by cubing both sides to eliminate the cube root:

\[
x^3 + 14x^2 + 49x + 20 = (x + 5)^3.
\]

Next, we expand the right side:

\[
(x + 5)^3 = x^3 + 15x^2 + 75x + 125.
\]

Now we can set the two sides equal to each other:

\[
x^3 + 14x^2 + 49x + 20 = x^3 + 15x^2 + 75x + 125.
\]

Subtract \(x^3\) from both sides:

\[
14x^2 + 49x + 20 = 15x^2 + 75x + 125.
\]

Now, let's move everything to one side of the equation:

\[
14x^2 + 49x + 20 - 15x^2 - 75x - 125 = 0.
\]

This simplifies to:

\[
-x^2 - 26x - 105 = 0.
\]

Multiplying by -1 gives:

\[
x^2 + 26x + 105 = 0.
\]

Now, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = 26\), and \(c = 105\).

Calculating the discriminant:

\[
b^2 - 4ac = 26^2 - 4 \times 1 \times 105 = 676 - 420 = 256.
\]

Now, applying the quadratic formula:

\[
x = \frac{-26 \pm \sqrt{256}}{2 \times 1} = \frac{-26 \pm 16}{2}.
\]

Calculating the two possible values for \(x\):

1. \(x = \frac{-26 + 16}{2} = \frac{-10}{2} = -5\)
2. \(x = \frac{-26 - 16}{2} = \frac{-42}{2} = -21\)

Thus, the solutions to the equation are:

\[
x = -5 \quad \text{and} \quad x = -21.
\]

Therefore, the correct response is:

**x = -21 and x = -5**.