Question

x2+x-(a+2)(a+1)=0 by quadratic equation by factorization method ​

Answer 1

Given:--

x^2+x-(a+2)(a+1)=0

=> D=b^2-4ac

=>(1)^2-4×(1)×-(a+2)(a+1)

=>1+4(a^2+3a+2)

for equal Roots...D=0

i.e;4a^2+12a+9=0

=>4a^2+6a+6a+9=0

=>2a(2a+3)+3(2a+3)=0

=>(2a+3)^2=0

=>a=-3/2.

Therefore,the required quadratic equation is--

x^2+x-(2-3/2)(1-3/2)=0

=>x^2+x+1/4=0

or 4x^2+4x+1=0.