Question

Find the nature of the roots of the equation in x $abx^{2}+(b^{2} -ac)x-bc=0$

Answer 1

Step-by-step explanation:

Given:-

The quadratic equation is abx^2 +(b^2-ac)x-bc =0

To find:-

Find the nature of the roots of the given equation

Solution:-

Given quadratic equation is

abx^2 +(b^2-ac)x-bc =0

On comparing with the standard quadratic equation ax^2+bx+c =0 then

a = ab

b= b^2-ac

c = -bc

To find the nature of the roots we have to find the value of the discriminant

We know that the standard quadratic equation ax^2+bx+c =0 then the discriminant is D = b^2-4ac

Now

D= (b^2-ac)^2 - 4(ab)(-bc)

=>D = b^4-2ab^2c +a^2c^2+4ab^2c

=>D= b^4+ 2ab^2c + a^2c^2

=>D = (b^2+ac)^2

Since the square of any integer is always positive

=>D> 0

Since the discriminant is greater than zero the roots of the given equation has real and distinct roots

Answer:-

The given quadratic equation has real and distinct roots

Used formulae:-

• the standard quadratic equation ax^2+bx+c =0 then the discriminant is D = b^2-4ac
• If D> 0 then the roots are distinct and real.
• if D<0 then the roots are imaginary(no real)
• If D = 0 then the roots are real and equal.