A triangle in a cubic equation?

Let $a$ , $b$ and $c$ be the sides of the triangle, and also the roots of the given equation. By Vieta's relations we have $s=\frac{a+b+c}{2}=-\frac{k}{2}$ It follows from Heron's formula that $A^2 = 6^2 = 36 = s(s-a)(s-b)(s-c)=-\frac{k}{2} \left( -\frac{k}{2} - a \right)\left( -\frac{k}{2} - b \right)\left( -\frac{k}{2} - c \right) = \frac{k}{2} \left( \frac{k}{2} + a \right)\left( \frac{k}{2} + b \right)\left( \frac{k}{2} + c \right)$ The last three factors suggest "shifting" the original polynomial by the substitution $x\rightarrow x -\frac{k}{2}$ , giving us the new polynomial $\left( x -\frac{k}{2} \right)^3 + k\left( x -\frac{k}{2} \right)^2 + 47\left( x -\frac{k}{2} \right) - 60=0$ with roots $a' = a + \frac{k}{2}$ , $b' = b + \frac{k}{2}$ , $c' = c + \frac{k}{2}$ . Therefore Vieta's relations again gives $36 = \frac{k}{2} \left( \frac{k}{2} + a \right)\left( \frac{k}{2} + b \right)\left( \frac{k}{2} + c \right) = \frac{k}{2}a'b'c' = \frac{k}{2}\cdot -\left[ \left( -\frac{k}{2} \right)^3 + k \left( -\frac{k}{2} \right)^2 + 47 \left( -\frac{k}{2} \right) - 60 \right]$ from which we get $\underline{k = -12}$ (Verify this one! :) ) Therefore, the original equation becomes $x^3 - 12x^2 + 47x - 60 = (x - 3)(x - 4)(x - 5) = 0$ from which $\underline{a=3,b=4,c=5}$ . Since $a^2 + b^2 = c^2$ , the triangle is a $\boxed{\text{right triangle}}$ , as desired. :))

Synopsis: I'll use a variation of the Heron's formula , $A = \dfrac14 \sqrt{(a^2+b^2+c^2)^2 - 2(a^4 + b^4 + c^4)}$ , where the expressions inside the bracket can be found via Vieta's Formula .

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Let $a,b,c$ denote the sides of the triangle in question, then they must be the zeros of the function $f(x) := x^3 + kx^2 + 47x - 60$ . Naturally, Vieta's formula tells us that its perimeter $a + b +c = -k > 0$ , so $k < 0$ .

We then have $a^2, b^2, c^2$ are the zeros of the function $f \left( \sqrt x \right) =x\sqrt x + k x + 47\sqrt x - 60$ .

By converting $f \left(\sqrt x \right) = 0$ into another cubic equation, we get

$\begin{aligned} x\sqrt x + k x + 47\sqrt x - 60 &=& 0 \\ x\sqrt x + 47\sqrt x &=& -kx + 60 \\ \left(x\sqrt x + 47\sqrt x \right)^2 &=& (-kx + 60)^2 \\ x^2 + 47^2 x + 94x^2 &=& k^2 x^2 - 120kx + 3600 \\ x^3 (1) + x^2 (94-k^2) + x(47^2 + 120k) - 3600 &=& 0 \end{aligned}$

Thus $a^2, b^2, c^2$ are the zeros of the function $g(x):= x^3 (1) + x^2 (94-k^2) + x(47^2 + 120k) - 3600$ .

By Vieta's formula, $\begin{cases} a^2 + b^2 + c^2 = k^2 -94 \\ a^2 b^2 + a^2 c^2 + b^2 c^2 = 47^2 + 120k \end{cases} \quad \Rightarrow \quad a^4 + b^4 + c^4 = (k^2 -94)^2 - 2(47^2 + 120k ) =k^4 - 188 k^2 - 240 k + 4418.$

Heron's formula tells us that $A = \dfrac14 \sqrt{(a^2+b^2+c^2)^2 - 2(a^4 + b^4 + c^4)}$ .

Given that $A = 6$ , then the equation above simplifies to $576 = -k^4 + 188 k^2 + 480 k$ . Rational root theorem shows that $k = -12$ is the only rational root. Factorizing $(k+12)$ out gives

$(k+12) (k^3 - 12 k^2 - 44 k + 48) = 0 .$

For completeness, let's prove that $k = -12$ is the only solution.

Notice from above that the sum of positive numbers $a^2 + b^2 + c^2 = k^2 - 94 > 0 \quad \Rightarrow \quad k^2 > 94 .$

Using intermediate value theorem and Descartes' rule of signs , we can verify that $h(k) := k^3 - 12 k^2 - 44 k + 48$ only has 1 negative root, which is in the interval $(-4,-3)$ . But since $k^2 > 94$ as well, $-4<k<-3$ is not possible. Hence, $k =-12$ is the only solution.

Substituting $k=-12$ back gives $f(x) = x^3 -12x^2 + 47x - 60 = (x-3)(x-4)(x-5)$ . Hence, the sides of the triangle in question are $3,4,5$ . Since $3^2 + 4^2 = 5^2$ . The triangle is $\boxed{\text{right-angled}}$ .

Let the roots be $a$ , $b$ and $c$ . Since these are the side lengths of a triangle, $a$ , $b$ , $c$ are all positive.

Expanding on Niranjan Khanderia's answer, by Vieta's formulas $abc$ is equal to the last coefficient divided by the first coefficient, provided that the polynomial is in the form $a_nx^2 + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ . In this case, $abc = \frac{60}{1} = 60$ .

Let's assume that there are integer $a$ , $b$ and $c$ . Factoring $60$ gives $2^2 \cdot 3 \cdot 5$ , and we have three choices for $(a,b,c)$ : $(4, 3, 5)$ , $(4, 6, 5)$ or $(2, 3, 10)$ . Of these three, only the triplet $3, 4, 5$ gives an $x$ -coefficient of $47$ .

By the answer options all of the triangles are either obtuse, acute or right. Since we have found a right-angled triangle, then all of the triangles with sides that are the roots of the equation $x^3+kx+47x-60$ are right .

I knew that 3-4-5 triangle has an area of 6. 3 4 5=60. And 3 4+4 5+5*3=47. These are also values we get from Vieta's formula.
3-4-5 is a rt. triangle.

3,4,5. Sided