Question

x (x-1)=1 quadratic formula
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Answer 1

Step-by-step explanation:

$x(x - 1) = 1 \\ \\ {x}^{2} - x = 1 \\ \\ it \: can \: be \: rewritten \: as \\ \\ {x}^{2} - x - 1 = 0 \\ general \: quadratic \: formula \: \\ is \: a {x}^{2} + bx + c$

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Answer 2

Explanation :-

$\implies\tt{x(x - 1) = 1}$

$\implies\tt \: x {}^{2} - x = 1$

$\implies\tt \: x {}^{2} - x - 1 = 0$

$\tt \: \: On \: \: comparing \: \: with \: \: \red{ax {}^{2} + bx + c = 0}\: \: we \: \: get,$

$\implies \tt \: a \: = 1 \: \: ,b = - 1 \: \: ,c = - 1$

As we know that,

$\dag \: \tt \: { \boxed{ \tt{ \red{x = \frac{ - b±\sqrt{b {}^{2} - 4ac } }{2a} }}}}$

[ Put the values ]

$\implies \: \tt \: x = \frac{ \: - ( - 1)±\sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 1) } }{2 \times 1}$

$\implies \tt \: x = \frac{1±\sqrt{1 + 4} }{2}$

$\implies \tt \: x = \frac{1± \sqrt{5} }{2}$

${ \boxed{ \blue{\tt \: \huge \: x = \frac{ 1 - \sqrt{5} }{2} \: \: \: \: OR \: \: \: \: \: \: x \: = \frac{1 + \sqrt{5} }{2} }}}$