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answers:
MyRank say: f(x) = 3x⁴ + 4x³ + 6(1-2)x² - 12ax + a²

→ 3x⁴ + 4x³ + 6(a-2)x² - 12ax + a²

f’(x) = 12x³ + 12x² + 12(a-2)x – 12a

f’’(x) = 36x² + 24x + 12 (a-2) = 0

f’’(x) = 0

∴ D = 0 → (24)² - 4.36.12(1-2) = 0

12(12)² - 4.36(a-2) = 0

(12)² - 4.3 (a-2) = 0

144 – 12 (a-2) = 0

9 – (a-2) = 0

a = 11.
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alex say: F(x) = 7x^3+6(a-2)x^2-12ax+a^2
f'(x) = 21x^2+12(a-2)x-12a = 0
or
7x^2+4(a-2)x-4a=0
for 1 solution ---> discriminant = 4(a-2)^2+28a=0
or
a^2+3a+4=0 ---> no solution
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llaffer say: Let's see what I get with the new function.

You have:

f(x) = 3x⁴ + 4x³ + 6(a - 2)x² - 12ax + a²

Stationary points happen when the first derivative is zero, so let's start there:

f'(x) = 12x³ + 12x² + 12(a - 2)x - 12a
0 = 12x³ + 12x² + 12(a - 2)x - 12a

We can divide both sides by 12 to simplify:

0 = x³ + x² + (a - 2)x - a

One thing that I noticed is that no matter what "a" is, x = 1 is always a root. To show this, if we substitute 1 for x, we get:

0 = 1³ + 1² + (a - 2)(1) - a
0 = 1 + 1 + (a - 2) - a
0 = 2 + a - 2 - a
0 = 0 + 0a
0 = 0
TRUE

So if x = 1 is always a root, then (x - 1) is always a factor, so let's divide:

. . . . _x²_+_2x_+_a_______
x - 1 ) x³ + x² + (a - 2)x - a
. . . . . x³ - x²
. . . . . -----------
. . . . . . . . . 2x² + (a - 2)x - a
. . . . . . . . . 2x² - 2x
. . . . . . . . .--------------
. . . . . . . . . . . (a - 2 + 2)x - a
. . . . . . . . . . . (a + 0)x - a
. . . . . . . . . . . ax - a
. . . . . . . . . . . ax - a
. . . . . . . . . . -------------
. . . . . . . . . . . . . . . 0

So now our equation is:

0 = x³ + x² + (a - 2)x - a
0 = (x - 1)(x² + 2x + a)

In order to have a single stationary point, we need one real zero. We already have that with x = 1. So we need the remaining quadratic to have complex roots (non-real). So now we can go to the discriminant and set up an inequality to check for values of "a" that result in a negative number:

b² - 4ac < 0

Using:

a = 1, b = 2, c = a:

2² - 4(1)a < 0
4 - 4a < 0
-4a < -4
a > 1

So if a is any value larger than 1, there will be one real root to the cubic which means that there is one stationary point in the original quartic equation.
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az_lender say: You probably miscopied the question. Why would they say 3x^3 + 4x^3 rather than just saying 7x^3. Oh well, I'll assume you coped it right.

Then F'(x) = 21x^2 + 12(a - 2)x - 12a.
You are seeking a value of "a" that would make the two roots of this quadratic function equal.

Hence, the discriminant should be zero, i.e.,
144(a^2 - 4a + 4) + 4*12*21a = 0 =>
36 a^2 - 144a + 144 + 12*21a = 0 =>
a^2 - 4a + 4 + 7a = 0 =>
a^2 + 3a + 4 = 0.

It doesn't work, as the roots are complex conjugates. Guess I must assume that you copied the question wrong in the first place.
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Doc say: Perhaps you could ask your teacher. None of us on here can help.
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vrfdp say: nsointas
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