Question

Good Question?

Predict the nature of the root of the following quadratic equation
$\orange{ \sf {x}^{2} -ax- {b}^{2} = 0 }$
​

Answer 1

Topic :-

Quadratic Equation

Given :-

x² - ax - b² = 0, a and b are real numbers.

To Find :-

Nature of roots.

Concept Used :-

Nature of Roots

Quadratic Equation : Ax² + Bx + C = 0

The expression B² - 4AC ≡ D is called the discriminant of the Quadratic Equation.

If D > 0 then roots are real and distinct.

If D = 0 then roots are real and coincident (equal).

If D < 0 then roots are imaginary.

Solution :-

General form of Quadratic Equation : Ax² + Bx + C = 0

Given Quadratic Equation : x² - ax - b² = 0

On comparing, we get,

A = 1

B = -a

C = -b²

Calculating value of Discriminant,

B² - 4AC

Substituting values,

(-a)² - 4(1)(-b²)

a² + 4b²

Now, we know that,

(f(x))² ∈ [0, ∞)

(Square of a function is always greater than or equal to 0.)

So,

a² and b² are non-negative numbers.

Thus,

a² + 4b² will be also a non-negative number.

(Non-negative number means All Positive Numbers including Zero.)

We can conclude that,

a² + 4b² ≥ 0

So,

If a² + 4b² > 0

then roots will be real and distinct and

If a² + 4b² = 0

then roots will be real and coincident (equal).

This case is only possible when a = b = 0.

Answer :-

So, the nature of roots of the equation x² - ax - b² = 0 will be :-

Real and distinct if a² + 4b² > 0 and

Real and equal if a² + 4b² = 0 or a = b = 0.