Question

( 2 x^{2}+40 x-5=0=9 )
Replace ( x^{prime} ) by ( 2 x-3=int x=frac{3+x}{2} )
( y=left(2(x-3)^{2}+x(2 x-3)-8=0right. )
( Rightarrow 8 x^{2}-24 x+18+8 x-17=0 )
( Rightarrow 8 x^{2}-16 x+ )
( 2left(frac{3+x^{1}}{2}right)^{2}+4left(frac{3+x^{1}}{2}right)-5=0 )
( Rightarrow frac{x^{2}+6 x^{1}+9}{2}+6+2 x^{2}-5=0 )
( Rightarrow quad x^{2}+10 x^{prime}+11=0 )
(2) ef rian

# 0.28 If aB are the roots of the equation 2x + 4x - 5 = 0, the equation whose roots are the reciprocals of 2a - 3 and 2B - 3 is- (A) x2 + 10x - 11 = 0 (B) 11x + 10x + 1 = 0 (C) x + 10x + 11 = 0 (D) 11x - 10x + 1 = 0

Solution