+118703
let r and s be the root of the quadratic x^2 + bx + c where b and c are constants. If (r-1)(s-1)=7, find b+c
the roots of 1x^2+bx+c are when
x^2 + bx + c = 0
x=−b±√b2−4∗1∗c2∗1x=−b±√b2−4c2r=−b+√b2−4c2s=−b−√b2−4c2 r=−b+√b2−4c2s=−b−√b2−4c2 (r−1)(s−1)=7rs−s−r+1=7rs−(s+r)=6 rs=b2−(b2−4c)4=c s+r=−b rs−(s+r)=6c−−b=6c+b=6
You should check this answer.
LaTex
x = {-b \pm \sqrt{b^2-4*1*c} \over 2*1}\\
x = {-b \pm \sqrt{b^2-4c} \over 2}\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\
~\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\~\\
(r-1)(s-1)=7\\
rs-s-r+1=7\\
rs-(s+r)=6\\~\\
rs=\frac{b^2-(b^2-4c)}{4}=c\\~\\
s+r=-b\\~\\
rs-(s+r)=6\\
c--b=6\\
c+b=6
