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X^2-(2b-1)x+(b^2-b-20)=0 |
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Answer» Please find below the solution to asked query: x2 − (2b − 1) x + (b2 − B − 20) = 0 Comparing the above equation with Ax2 + Bx + C = 0, We get, A = 1 B = −(2b − 1) C = b2 − b − 20 Now, D = B2 − 4AC = (2b − 1)2 − 4 (b2 − b − 20) = 4b2 + 1 − 4b − 4b2 + 4b + 80 = 81 > 0So, 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 Hope this would clear your doubt. |
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