Question

Determine the nature of the roots of the following quadratic equations:

(i) \((x-2a)(x-2b)=4ab\) 

(ii) \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\) 

(iii) \(2(a^2+b^2)x^2+2(a+b)x+1=0\)

(iv) \((b+c)x^2-(a+b+c)x+a=0\)

Answer 1

(i) \((x-2a)(x-2b)=4ab\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\((x-2a)(x-2b)=4ab\)

⇒ x² – (2a + 2b)x + 4ab = 4ab 

⇒ x² – (2a + 2b)x = 0 

D = (2a + 2b)² – 0 = (2a + 2b)²

Roots are real and distinct

(ii) \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\)

⇒ D = 576a²b²c²d² – 4 × 16 × 9 × a²b²c²d² = 0

Roots are real and equal

(iii) \(2(a^2+b^2)x^2+2(a+b)x+1=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(2(a^2+b^2)x^2+2(a+b)x+1=0\)

⇒ D = 4(a + b)² – 4 × 2 × (a² + b²)

⇒ D = 4(a + b)² – 4 × 2 × (a² + b²)

Roots are not real

(iv) \((b+c)x^2-(a+b+c)x+a=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\((b+c)x^2-(a+b+c)x+a=0\)

⇒ D = (a + b + c)² – 4a(b + c)

⇒ D = a² + b² + c² – 2ab – 2ac + 2bc

⇒ D = (a – b – c)²

Thus, roots are real and unequal