If `a + b != 0` then the roots of `(a + b) x^2 + ax - b = 0` are going to be real. Prove it.

The roots of the equation x^2+ax+b=0 are

If the equaion x^(2) + ax+ b = 0 has distinct real roots and x^(2) + a|x| +b = 0 has only one real root, then

If the equaion x^(2) + ax+ b = 0 has distinct real roots and x^(2) + a|x| +b = 0 has only one real root, then

The roots of the equation x^(2) + ax + b = 0 are "______" .

If a and b are the roots of the equation x^(2) + ax + b = 0, a ne 0, b ne = 0 , then the values of a and b are respectively

The roots of the equation ax+(a+b)x-b*=0 are

If a and b are the roots of the equation x^2+ax+b=0 ,then a+b=

If a and b are the roots of the equaltion x^(2)+ax-b=0 , then find a and b.

If a and b ( != 0) are roots of x^(2) +ax + b = 0 , then the least value of x^(2) +ax + b ( x in R) is :

If a and b(!=0) are the roots of the equation x^2+ax+b=0 then the least value of x^2+ax+b is