Question

(x-2a)(x-2b)=4ab
find the nature of the roots

Answer 1

Step-by-step explanation:

Given equation is:-

(

$(x - 2a)(x - 2b) = 4ab \\ {x}^{2} - 2bx - 2ax + 2ab - 4ab = 0 \\ {x}^{2} - (2b + 2a)x - 2ab = 0 \\ {b}^{2} - 4ac = ( - (2b + 2a)) {}^{2} - 4(1)( - 2ab) \\ = 4 {b}^{2} + 8ab + 4 {a}^{2} + 8ab \\ = 4 {a}^{2} + 16ab + 4 {b}^{2} \\ = 4(a {}^{2} + 4ab + {b}^{2} ) \\ > 0$

As discriminant b^2 - 4ac is greater than zero,the roots are real and unequal.

Answer 2

Answer:

$\large \purple{Answer}$➙

Nature of rootsↆ

(x−2a)(x−2b)=4ab

⇒ x 2 −2bx−2ax+4ab=4ab

⇒ x 2 −2(a+b)x=0

⇒ On comparing with,

##ax^2+bx+c=0$$

⇒ We get, a=1,b=−2(a+b),c=0

⇒ Discriminant of the quadratic,equation =b 2 −4ac

⇒ [−(a+b)] 2 −4×1×0

⇒ (a+b) 2

∴ b 2 −4ac>0

Thus, roots are real and distinct.

$\large \orange{Mabelrose♡︎}$