solving quartic equations

[DDuino]

does anybody know how to solve (x^4+x^3+x^2+x+1)=0?

[HelloooG]

X1= 0.31.....

X2 = 0.31....

X3 = -0.81....

X4 = -0.81.....

(Poly.SMLT - Ti-84-Plus)

[DDuino]

hmm...which formula did you use man?just the calculator ?

[HelloooG]

Yeapp

[Slovakov]

I'm not sure whether the app poly.smlt is allowed during the exam. At least we obtain "clear" GDCs with no apps, and can not have our own calculators during the exam.

This can be solved by putting your formula as a function, plotting a graph and checking its zeros. They are the real roots of this polynomial. Don't know whether the TI can show the complex ones.

Edited April 28, 2011 by Slovakov

[HelloooG]

Yes the programm is allowed during the exams... They don't delete the originals programs

[genepeer]

This can be solved by putting your formula as a function, plotting a graph and checking its zeros. They are the real roots of this polynomial. Don't know whether the TI can show the complex ones.

That's the expected way of solving it anyway. They wouldn't give an equation with complex solutions unless it's easily factorisable, in which case, you won't even need a calc.

Btw, those answers are wrong

y=x⁴+x³+x²+x+1 has no real roots. It never crosses the x-axis. Solving it:

x⁴+x³+x²+x+1 = (x⁵-1)/(x-1) : Geometric series, and x not equal to 1 (otherwise, division by zero)

so:

(x⁵-1)/(x-1) = 0

x⁵-1 = 0

x⁵ = 1 = e^{i(0+k 2π)}

x = e^{i(k 2π)/5} for k=0, 1, 2, 3, 4. Five solutions for fifth root. Remember, N-solutions for n-th root. Now substitute the k's...

= 1, e^i(2π/5), e^i(4π/5), e^i(6π/5), e^i(8π/5)

Remember that x can't be 1 due to division by 0 mentioned earlier, and it's obviously not a solution to the given equation, x⁴+x³+x²+x+1=0

Edited April 28, 2011 by Aboo

[DDuino]

This can be solved by putting your formula as a function, plotting a graph and checking its zeros. They are the real roots of this polynomial. Don't know whether the TI can show the complex ones.

That's the expected way of solving it anyway. They wouldn't give an equation with complex solutions unless it's easily factorisable, in which case, you won't even need a calc.

Btw, those answers are wrong

y=x⁴+x³+x²+x+1 has no real roots. It never crosses the x-axis. Solving it:

x⁴+x³+x²+x+1 = (x⁵-1)/(x-1) : Geometric series, and x not equal to 1 (otherwise, division by zero)

so:

(x⁵-1)/(x-1) = 0

x⁵-1 = 0

x⁵ = 1 = e^{i(0+k 2π)}

x = e^{i(k 2π)/5} for k=0, 1, 2, 3, 4. Five solutions for fifth root. Remember, N-solutions for n-th root. Now substitute the k's...

= 1, e^i(2π/5), e^i(4π/5), e^i(6π/5), e^i(8π/5)

Remember that x can't be 1 due to division by 0 mentioned earlier, and it's obviously not a solution to the given equation, x⁴+x³+x²+x+1=0

so basically...we see it as a x5-1=0...right?and then you just divided it with (x-1)...???

[genepeer]

so basically...we see it as a x5-1=0...right?and then you just divided it with (x-1)...???

Not being rude or anything but I don't get what you're asking?! If there's a step you don't understand, I'd explain it.

[DDuino]

y=x4+x3+x2+x+1 has no real roots. It never crosses the x-axis. Solving it:

x4+x3+x2+x+1 = (x5-1)/(x-1) : Geometric series, and x not equal to 1 (otherwise, division by zero)

dont get this bit...thanks for help

[genepeer]

y=x⁴+x³+x²+x+1 has no real roots. It never crosses the x-axis. Solving it:

The equation is x⁴+x³+x²+x+1=0.

We look at the LHS (left-hand side) and RHS as two different lines on a graph, and the equation is asking for the intersection points since it's equating the two lines.

Line 1: y=x⁴+x³+x²+x+1

Line 2: y=0 : this is also the x-axis.

If you have a GDC or graphing software, you'll see that these two lines will never meet/cross/intersect. This means that there's is no real x that satisfies the equation, otherwise we'd have an intersection. However, all polynomial equations have solutions, so if it's not real (no intersection), it has to be complex. Hence why 1 was cancelled out because it's a real number (and the division by zero too).

x⁴+x³+x²+x+1 = (x⁵-1)/(x-1) : Geometric series, and x not equal to 1 (otherwise, division by zero)

Geometric series:

with a=1, r=x, and n=5 to get the (1-x⁵)/(1-x)... multiply top & bottom by -1 and you get (x⁵-1)/(x-1)

The next step was just multiplying both side by (x-1) and the "+k 2π" is because angles are periodic every after 2π (or 360^o e.g. 20^o=380^o=740^o=...)