+0  
 
+1
109
5
avatar+309 

Let j and k be the solutions to \(7x^2+x-5=0\). Compute \((j-4)(k-4)\).

 Apr 2, 2020
edited by mathmathj28  Apr 2, 2020
 #1
avatar
+1

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.

 Apr 2, 2020
 #2
avatar
+1

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))

 Apr 2, 2020
 #3
avatar
+1

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=\(rsq=-\frac{d}{a}\)

Guest Apr 2, 2020
 #4
avatar
+1

Oops, the calculation is 15.857 (not 15.32)

Guest Apr 2, 2020
 #5
avatar+309 
+1

Thanks so much!!

 Apr 2, 2020

7 Online Users

avatar
avatar