If αand βare the roots of the equation ax2+bx+c=0(c≠0), then the equation, whose roots are1/[aα+b]and 1/[aβ+b], is
acx2–bx+1=0
x2–acx+bc+1=0
acx2+bx–1=0
x2+acx+11=0
Explanation for the correct option:
Find the equation :
Given αand βare the roots of the equation ax2+bx+c=0(c≠0),
⇒ɑ+β=–b/a,ɑβ=c/a
The required equation is
x2–1(aα+b)+1(aβ+b)x+1(aα+b)×1(aβ+b)=0⇒x2–(a(ɑ+β)+2b)(a2ɑβ+ab(ɑ+β)+b2)x+1(a2ɑβ+ab(ɑ+β)+b2)=0⇒x2–a-ba+2b)(a2ca+ab-ba+b2)x+1(a2ca+ab-ba+b2)=0⇒x2–bacx+1ac=0⇒acx2–bx+1=0
Hence the correct option is A.
If α,β are the roots of the equation ax2+bx+c=0 then the equation whose roots are α+1β and β+1α is