show that the roots of equation (x-a)(x-b)=abx^{2 ;} a,b belongs to real number

show that when they are equal?

please explain it i m too weak in quardatic

Question

show that the roots of equation (x-a)(x-b)=abx2 ;
a,b belongs to real number
show that when they are equal?
please explain it i m too weak in quardatic

Pronay · Accepted Answer

$(x - a) (x - b) = a b x^{2} \Rightarrow (x^{2} - (a + b) x + a b) = {a b x}^{2} \Rightarrow (a b - 1) x^{2} + (a + b) x - a b = 0$

D of the above equation is

${(a + b)}^{2} + 4 a b (a b - 1) = {a^{2} + 2 a b + b}^{2} - 4 a b + 4 {(a b)}^{2} = {(a - b)}^{2} + 4 {(a b)}^{2} > 0$

Therefore roots of given equation i.e. a and b belongs to real numbers

Both are equal if $a b = 1$

show that the roots of equation (x-a)(x-b)=abx2 ; a,b belongs to real number show that when they are equal? please explain it i m too weak in quardatic

show that the roots of equation (x-a)(x-b)=abx^{2 ;} a,b belongs to real number

show that when they are equal?

please explain it i m too weak in quardatic