Question

find the nature of roots of the equation x^2+ax-4​

Answer 1

Answer:

The nature of the roots of a quadratic equation depends entirely on the value of its discriminant b2 - 4ac.

In a quadratic equation ax2 + bx + c = 0, a ≠ 0 the coefficients a, b and c are real. We know, the roots (solution) of the equation ax2 + bx + c = 0 are given by x = −b±b2−4ac√2a.

1. If b2 - 4ac = 0 then the roots will be x = −b±02a = −b−02a, −b+02a = −b2a, −b2a.

Clearly, −b2a is a real number because b and a are real.

Thus, the roots of the equation ax2 + bx + c = 0 are real and equal if b2 – 4ac = 0.

Answer 2

The nature of the roots of a quadratic equation depends entirely on the value of its discriminant b2 - 4ac.

In a quadratic equation ax2 + bx + c = 0, a ≠ 0 the coefficients a, b and c are real. We know, the roots of the equation ax2 + bx + c = 0 are given by x = −b±b2−4ac√2a.

For example▶

1. If b2 - 4ac = 0 then the roots will be x = −b±02a = −b−02a, −b+02a = −b2a, −b2a.

Clearly, −b2a is a real number because b and a are real.

Thus, the roots of the equation ax2 + bx + c = 0 are real and equal if b2 – 4ac = 0.

Hope it will help you...☺