Let \(a,b,c\) be the real roots of \(x^3 -4x^2 -32x+17=0.\) Solve for \(x\) in \(\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.\)
Please help!
+15001 Let a, b, c be the real roots of \(x^3 -4x^2 -32x+17=0.\)
Solve for x in: \(\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.\)
Hello Guest!
WolframAlpha calculated:
a = - 4.3195
b = 0.50355
c = 7.8159
a + b + c = 4
abc = - 17
\(\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0\\ \sqrt[3]{x +4.3195} + \sqrt[3]{x - 0.50355} + \sqrt[3]{x - 7.8159} = 0\)
No solution found.
!
+118690 Hi asinus,
That was good thinking.
But I graphed your equation using Desmos and there was a solution
Hello Guest!
If a,b,c are the solutions of the polynomial equation: \(x^3-4x^2-32x+17=(x-a)(x-b)(x-c)=0\)
Then, by Vietas' formulae: \(a+b+c=4\)
\(ab+ac+bc=-32\)
\(abc=17\)
Using the following identity:
\(y_1^3+y_2^3+y_3^3-3y_1y_2y_3=(y_1+y_2+y_3)(y_1^2+y_2^2+y_3^2-y_1y_2-y_1y_3-y_2y_3)\) (*)
(More commonly written as: \(x^3+y^3+z^3=3xyz\) if and only if \(x+y+z=0\)).
Then:
Let \(y_1=\sqrt[3]{x-a}\) , \(y_2=\sqrt[3]{x-b}\) , \(y_3=\sqrt[3]{x-c}\)
We are given: \(y_1+y_2+y_3=0\)
Hence, (*) becomes:
\((x-a)+(x-b)+(x-c)=3\sqrt[3]{(x-a)(x-b)(x-c)}\)
(Notice: \((x-a)(x-b)(x-c)=x^3-4x^2-32x+17\))
Thus,
\( 3x-(a+b+c) = 3\sqrt[3]{x^3-4x^2-32x+17}\)
(By Vietas formula: a+b+c=4, substitute and cube both sides)
\(\iff (3x-4)^3=27(x^3-4x^2-32x+17)\)
Expanding:
\(\iff 27x^3-108x^2+144x-64=27x^3-108x^2-864x+459\)
Simplify to get a linear equation:
\(144x-64=-864x+459 \implies 1008x=523 \implies x=\frac{523}{1008}\)
Therefore, \(x=\dfrac{523}{1008}\)
Hi asinus, by the way, Wolframalpha approximated a,b,c a lot.
Because when using mathway.com, it gives the following values:
\(a=-4.31946756\)
\(b=0.5035452\)
\(c=7.81592235\)
So:
\(\sqrt[3]{(x+4.31946756)}+\sqrt[3]{(x-0.5035452)}+\sqrt[3]{(x-7.81592235)}=0\)
And, then graphing this equation as Melody did (But using the more accurate values of a,b,c):
Giving the approximation: \(0.519\) (And the exact solution is: \(\frac{523}{1008}=0.518849206..\)).
Hope this helps!
