+306 Let j and k be the solutions to \(7x^2+x-5=0\). Compute \((j-4)(k-4)\).
The solutions to 7x^2 + x - 5 = 0 are (-1 + i*sqrt(139))/14 and (-1 - i*sqrt(139))/14. Then ((-1 + i*sqrt(139))/14 - 4)((-1 - i*sqrt(139))/14) - 4) = 111/7.
j and k are solutions means they are the roots of the equations
Relation of roots to the polynomial (Quadratic equation in this case) is as follows:
j+k= \(-\frac{1}{7}\) (1)
jk=\(-\frac{5}{7}\) (2)
\(jk-4j-4k+16\) (This is the expansion of (j-4)(k-4)
\(jk-4(j+k)+16\) (Factored -4)
\(-\frac{5}{7}-4(-\frac{1}{7})+16=15.32\) (Substitute (1), (2))
General rule:
For example: \(ax^2+bx+c=0\)
Sum of roots: \(-\frac{b}{a}\)
Product of roots: \(\frac{c}{a}\)
This stands for every polynomial and not restricted to a second degree only.
For instance,
\(ax^3+bx^2+cx+d=0\) (Where r,s,q are the roots)
Sum of roots: \(-\frac{b}{a}\)
Sum of each pair of roots: \(rs+sq+rq=\frac{c}{a}\)
Product of roots=\(rsq=-\frac{d}{a}\)
