Question

..

x² - (2b - 1)x + (b² - b - 20) = 0

Factorise.⇄

Answer 1

Answer:

x² - (2b - 1)x + (b² - b - 20) = 0

x² - (2b - 1)x + [ b² - 5b + 4b - 20] = 0

x² - (2b - 1)x + [ b (b - 5) + 4 (b - 5)] = 0

x² - (2b - 1)x + (b - 5 )( b + 4 ) = 0

x² - (b - 5)x - (b + 4)x + (b - 5 )( b + 4 ) = 0

x [ x - (b -5)] - (b + 4)[ x - (b - 5)] = 0

[ x - (b -5) ] [ x - (b + 4) ] = 0

Therefore, using zero product rule;

x = (b - 5) or x = (b + 4)

__________________

Thanks for the question!

Answer 2

$\huge\bold\pink{Solution}$

$<b><u><font color=red>$

x2 − (2b − 1) x + (b2 − b − 20) = 0

Comparing the above equation with Ax2 + Bx + C = 0

We get,

A = 1B = −(2b − 1)C = b2 − b − 20

Now, D = B2 − 4AC= (2b − 1)2 − 4 (b2 − b − 20)= 4b2 + 1 − 4b − 4b2 + 4b + 80= 81 > 0

So, the above equation has 2 real and distinct roots given by,

x = −B + D√2A   

 x = −B − D√2ᴀ

⇒ x = 2b − 1 + 92    or     x = 2b − 1 − 92$<b>$⇒ x = b + 4    or    x = b − 5