solve : x^2-(2b-1)x+(b^2-b-20)
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Answer:
Step-by-step explanation:
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 or x = −B − D√2A⇒ x = 2b − 1 + 92 or x = 2b − 1 − 92⇒ x = b + 4 or x = b − 5
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