If the equation x^2+2 a x+a= a^2+x- 1 16 - 1 16 has no real roots, then the range of a is (p, q) where p and q are rational numbers. If p^2+q^2= c d (where c and d are co-prime) then find the value of 4 c^2-d^2 .

Step-by-step solution

aligned x^2+2 a x+a & = a^2+x- 1 16 - 1 16 (x+a)^2+a-a^2+ 1 16 & = (x+a)- (a-a^2+ 1 16 ) aligned Let x+a=y and a-a^2+ 1 16 =p Then y^2+p= y-p f(x)=f⁻¹(x) f(x)=x has no real root y^2+p=y has no real root array rr D 1 4 a-a^2+ 1 16 > 1 4 a ( 1 4 , 3 4 ) & c d = 5 8 4 c^2-d^2 = array c d = 5 8 4 c^2-d^2 = 100-64 =6