Inspired by Mehul Chaturvedi

$8x^4 - 24x^3 + 27x^2 = bx-c$ We want to find a line $y=bx-c$ that intercepts the curve $y = 8x^4 - 24x^3 + 27x^2$ at four points. Why would there only be one possible gradient of such a line? Could we not just tilt it very slightly? We are free to adjust the value of $c$ , the y-intercept, all we like. A little geometrical intuition tells us that this is only possible if it is a quadruple root. Indeed, differentiating $y = 8x^4 - 24x^3 + 27x^2$ gives a cubic with an infection at the point $x=3/4$ : that is, its own derivative is a quadratic with a double root at $x=3/4$ . This is the point where our line must touch. Put $x=3/4$ in the cubic we had to find the gradient of the quartic at that point (and thus the gradient of the tangent line $y=bx-c$ ) as 13.5.

Just plug a hole in Aditya's argument,where he relied on AM-RMS inequality and the extra condition that all roots are positive.

Instead of using AM-RMS inequality, we could use the following inequality

$3*\sum_{i=1}^4\alpha^2\geq 2*\sum_ {i=1}^4\alpha*\beta$

Which is true for all real numbers of $\alpha$ and $\beta$ , no need for them to be positive, the equality holds when all numbers are equal. Since both sides of this inequality equal $\frac{27}{4}$ in our case, we deduce that all roots are equal, and the rest just follows Aditya's reasoning.

Consider $g(x) = 2 f(x) - ( 2x-3/2) ^ 4 = (27 - 2b) x + (2c - \frac{ 81}{ 16} )$ .

When $f(x) = 0$ , this implies that $( 2x - 3/2) ^ 4 = - (27 - 2b) x - (2c - \frac{ 81}{ 16} )$ .

How can a perfect 4th power intersect a line at 4 points, even if we include multiplicity?

One way is for the line to be tangential to the vertex of the 4th power, which will give us 4 roots. In this case, we must have $- ( 27 - 2b) x - (2c - \frac{81}{16} ) = 0$ , or that $b = 13.5$ .

If the line is not tangential to the vertex of the 4th power, consider the following:
1. The line cannot intersect the 4th power at 4 or 3 points.
2. If the line intersects the 4th power at 2 points, then neither of these points are tangent, and hence the maximum number of real roots is 2.
3. If the line intersects the 4th power at 1 point that is not the vertex, then the number of multiplicities is at most 2. Hence the maximum number of real roots is 2.

Hence, the only possible scenario gives us $b = 13.5$ , and all of the roots are equal.

We count roots up to multiplicity.

For the equation to have 4 real roots, f'(x) should have 3 real roots. And for f'(x) to have 3 real roots, f''(x) should have 2 real roots.

$f'(x)=32x^3-72x^2+54x-b$

$f''(x)=96x^2-144x+54$ . Observe that this quadratic expression has a repeated root at $x=0.75$ .

This tells us that $f'(x) = A( x- 0.75)^3 + B$ , where $A, B$ are some constants. In order for $f'(x)$ to have 3 real roots, we must have $B= 0$ .

Then, with $A = 32$ , we can calculate that $b = 13.5$ .

It remains to verify that $f(x) = 0$ indeed has 4 real roots, and this is because $f(x) = 8 ( x - 0.75)^4$ .

Apparently, any change in $b$ or $c$ would cause some of the real roots to vanish. This is only possible if $f$ has a minimum or maximum at a root with multiplicity at least 3.

Therefore I have good hopes that this quartic is of the form $f(x) = a(x-r)^4.$ Of course we choose $a = 8$ and $r = \tfrac34$ : $8(x-\tfrac34)^4 = 8x^4-24x^3+27x^2-13\tfrac12x+\tfrac{81}{32}.$ Bingo! We get immediately $b = 13\tfrac12$ .

I was just simply trying out a few things to see if I can solve this, and I stumbled upon this solution:

For a quartic polynomial to have $4$ real roots, its derivative must have $3$ real roots ( Reference )

The derivative, $f'(x) = 32x^3 - 72x^2 + 54x - b$

The discriminant for the cubic derivative,

$\Delta = b^2c^2 -4ac^3 -4b^3d - 27a^2d^2 + 18abcd\\ =(-72)^2(54)^2 - 4(32)(54)^3 - 4(-72)^3(-b) - 27(32)^2(-b)^2 + 18(32)(-72)(54)(-b)\\ =15116544 - 20155392 -1492992b + 2239488b - 27648b^2\\ =-5038848 +746496b - 27648b^2\\ =-6912(4b^2-108b + 729)\\ =-6912(2b-27)^2$

For all values of $b$ :

$(2b-27)^2 \geq 0\\ -6912(2b-27)^2 \leq 0\\ \Delta \leq 0$

For a cubic polynomial to have $3$ real roots, $\Delta \geq 0$ . This implies that we have only one possible discriminant value

$\Delta = 0\\ -6912(2b-27)^2 =0\\ (2b-27)^2 = 0\\ 2b-27 = 0\\ b= \boxed{13.5}$

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Bonus:

You can then use the quartic polynomial discriminant to show that for $b=13.5$ , the polynomial either has $4$ equal real roots, $2$ real roots or no real roots at all (though I seriously do not recommend you to try this)

To understand this section, refer to this Wikipedia page

$\Delta = 4(32c-81)^3\\ P=0\\ Q=0\\ \Delta_0 = 3(32c-81)\\ D = 1024(32c- 81)$

For $c < \dfrac{81}{32}, \Delta < 0$ , $f(x)$ has $2$ real roots

For $f(x)$ to have $4$ real roots, $c \geq \dfrac{81}{32}$

If $c >\dfrac{81}{32}$ , then $\Delta > 0, D > 0$ , which implies that $f(x)$ has no real roots

Therefore, $c$ can only have one value: $\boxed{\dfrac{81}{32}}$

When $c=\dfrac{81}{32}, \Delta = 0,\; \Delta_0 = 0,\;D = 0$ , which implies that $f(x)$ has $4$ equal real roots: $x=-\dfrac{b}{4a} = \dfrac{3}{4}$

The polynomial is $f(x) = 8x^4 - 24x^3 + 27x^2 - \dfrac{27}{2} + \dfrac{81}{32} = \dfrac{1}{32}\left(4x-3\right)^4$

There's something missing in the question and it is that all roots are positive.

We now see that if roots are $\alpha, \beta,\gamma , \delta$ , then we have

$\sum \alpha = 3 , \sum \alpha \beta = \dfrac{27}{8}$ .

Here $\sum \alpha = \alpha + \beta + \gamma + \delta \\ \sum \alpha \beta=\alpha \beta + \beta \gamma + \delta\gamma +\alpha \delta + \beta \delta + \alpha \gamma$

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We know that $(\sum \alpha)^2 = \sum \alpha^2 + 2 \sum \alpha \beta$ , hence we have

$3^2 = \sum \alpha^2 + 2 \times \dfrac{27}{8}$

$\therefore \sum \alpha^2 = 9 - \dfrac{27}{4} = \dfrac{9}{4}$

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If the roots are all positive, by the A.M.-R.M.S inequality, we have

$\sqrt{\dfrac{\alpha^2 + \beta^2 + \gamma^2 + \delta^2}{4} } \geq \dfrac{\alpha + \beta + \gamma + \delta}{4}$

And equality occurs only when $\alpha =\beta= \gamma=\delta$ .

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From the question, $\sqrt{\dfrac{\alpha^2 + \beta^2 + \gamma^2 + \delta^2}{4} } = \sqrt{\dfrac{9/4}{4}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$

and $\dfrac{\alpha + \beta + \gamma + \delta}{4} = \dfrac{3}{4}$

AM=RMS $\implies$ $\alpha = \beta = \gamma = \delta= \dfrac{3}{4}$ .

This proves $f(x) = 8\times \bigl(x-\frac{3}{4} \bigl)^4$ , giving $\boxed{b=13.5}$

Let x = y + k and substitutes into the polynomial given,

With k = 3/ 4, terms of y^3 and y^2 becomes zero,

8 y^4 + (27/ 2 - b) y + [243/ 32 - (3/ 4) b + c]

Again, with b = 27/ 2,

8 y^4 + [225/ 32 + c]

Such y can easily be real with proper c.

Hence, x = y + 3/ 4 are also real.

So, b = 13.5 gives 4 real roots.

{My computing solver is faulty and doesn't give 4 real roots with these anyway.}