Question Number 53500 by Otchere Abdullai last updated on 22/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

Commented by Otchere Abdullai last updated on 22/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
![2)(x+ay+b)(x+cy+d)=x^2 +5xy+x+ny−n x^2 +cxy+xd+axy+acy^2 +ayd+bx+bcy+bd=x^2 +5xy+x+ny−n x^2 +xy(a+c)+x(b+d)+y(ad+bc)+acy^2 +bd=x^2 +5xy+x+ny−n so a+c=5 b+d=1 ad+bc=n ac=0 bd=−n ac=0 a(5−a)=0 either a=0 or a=5 when a=0 c=5 and when a=5 c=0 now let a=0 c.=5 ad+bc=n 0×d+b×5=n [b=(n/5)] a=0 b=(n/5) c=5 d=1−b=1−(n/5) bd=−n ((n/5))×(1−(n/5))=−n (1/5)×(((5−n)/5))=−1 5−n=−25 −n=−30 ...n=30](https://www.tinkutara.com/question/Q53508.png)
Commented by Otchere Abdullai last updated on 22/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
