We are given the polynomial:
2x³ - 3x² + 4x - 1 = 0
We need to find the value of r² + s² + t² where r,s,t are the solutions.
Simplifying r² + s² + t²:
r² + s² + t² = (r + s + t)² - 2(rs + st + rt)
[since a²+b²+c² = (a+b+c)² - 2(ab + bc + ca)]
Using the Vieta's formulae:
Above, we can see that in order to find the value of r² + s² + t², we need to find the value of the sum of solutions and sum of solutions taken 2 at a time.
Sum of roots:
We know that the Vieta's formula to find the sum of roots is:
Sum of roots = [tex]-\frac{b}{a}[/tex]
Where b and a are the coefficients from the general form: ax³ + bx² + cx - d
We notice that for the given equation, b = -3 ; a = 2
Sum of roots = [tex]-(\frac{-3}{2})[/tex]
Sum of roots = [tex]\frac{3}{2}[/tex]
Hence, the sum of roots (r + s + t) = 3/2
Sum of roots taken two at a time:
We know that the Vieta's formula to find the sum of roots taken 2 at a time is: Sum of roots taken 2 at a time = [tex]\frac{c}{a}[/tex]
Where c and a are the coefficients from the general form: ax³ + bx² + cx - d
for the given equation, c = 4 ; a = 2
Sum of roots taken 2 at a time = [tex]\frac{4}{2}[/tex]
Sum of roots taken 2 at a time = 2
Hence, the sum of roots taken 2 at a time (rs + st + rt) = 2
Finding the value of r² + s² + t²:
We proved before that:
r² + s² + t² = (r + s + t)² - 2(rs + st + rt)
now that we know the values of (r + s + t) and (rs + st + rt), we can plug those values to get our answer
r² + s² + t² = ([tex]\frac{3}{2}[/tex])² - 2(2)
r² + s² + t² = ([tex]\frac{9}{4\\}[/tex]) - 4
r² + s² + t² = [tex]\frac{9 - 16}{4\\}[/tex]
r² + s² + t² = [tex]\frac{-7}{4}[/tex] = -1.75
Hence, the value of r² + s² + t² is -1.75
