Question

9) Solve

(x² + 3x)³-16(x² + 3x) - 36 = 0
for x by using substitution.
a) x = -3,1,2,6
b) x = -4,0,1,5
c) x = -5,-1,0,4
d) x = -6, -2,-1,3
e) x = -7,-3, -2,2

Answer 1

Answer: 0

Explanation:

To solve the equation (x² + 3x)³ - 16(x² + 3x) - 36 = 0 using substitution, let's make a substitution:

Let u = x² + 3x.

Now, we can rewrite the equation in terms of u:

u³ - 16u - 36 = 0.

Let's solve this equation for u by factoring:

(u - 6)(u² + 6u + 6) = 0.

Now, we have two possible cases:

Case 1: u - 6 = 0.

This gives us u = 6.

Case 2: u² + 6u + 6 = 0.

To solve this quadratic equation, we can use the quadratic formula:

u = (-b ± √(b² - 4ac)) / (2a),

where a = 1, b = 6, and c = 6.

Plugging in these values, we get:

u = (-6 ± √(6² - 4(1)(6))) / (2(1)),

u = (-6 ± √(36 - 24)) / 2,

u = (-6 ± √12) / 2,

u = (-6 ± 2√3) / 2,

u = -3 ± √3.

Now that we have the possible values of u, let's substitute back to find the corresponding values of x:

Case 1: u = 6.

Since u = x² + 3x, we have x² + 3x = 6.

Rearranging, we get x² + 3x - 6 = 0.

Hope it helps!