Question

x²-(2b-1)x+(b²-b-20)=0​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

$so \: here \: given \: quadratic \: equation \: is \\ x {}^{2} - (2b - 1)x + (b {}^{2} - b - 20) = 0 \\ so \: comparing \: the \: given \: quadratic \: equation \: with \: ax {}^{2} + bx + c = 0 \\ we \: get \\ a = 1 \\ b = - (2b - 1) \\ c = (b {}^{2} - b - 20)$

$so \: now \: using \: \\ Δ = b {}^{2} - 4ac \\ = ( - (2b - 1)) {}^{2} - 4(b {}^{2} - b - 20) \times 1 = 0 \\ = (2b - 1) {}^{2} - 4b {}^{2} + 4b + 80 \\ = 4b {}^{2} + 1 - 4b - 4b {}^{2} + 4b + 80 \\ = 80 + 1 \\ Δ = 81$

$now \: by \: using \: formula \: method \: \\ we \: get \\ x = - b \: ± \: \sqrt{b {}^{2} - 4ac } / \: 2a \\ = - ( - (2b - 1)) \: ± \sqrt{81} \: / \: (2 \times 1) \\ = 2b - 1 \: ± \: 9\: /2 \\ ie \: \: x = 2b - 1 + 9 \: / \: 2 \\ or \: \\ x = 2b - 1 - 9 \: / 2 \\ x = 2b+ 8 \: / \: 2 \\ or \\ x = 2b - 10 \: / \: 2$

$ie \: \: x = 2(b + 4) \: / \: 2 \\ or \\ x = 2(b - 5) \: / \: 2 \\ so \: hence \: \\ x = b + 4 \: \: or \: \: x = b - 5$