Question

solve the quadratic equation by shreedhracharya's rule x²-(√3+1)+√3=0​

Answer 1

Answer:

$there \: are \: two \: root \: of \: x \\ in \: quadratic \: equation \\ x1 = \frac{ - b + \sqrt{d} }{2a} \\ x2 = \frac{ - b - \sqrt{d} }{2a} \\ where \: d \: is \: discriminant \: \\ = \sqrt{ {b}^{2} - 4ac} \\ a = coeffient \: of \: {x}^{2} \\ b = coeffient \: of \: x \\ c = constant \: term$

Step-by-step explanation:

$a = 1 \\ b = - ( \sqrt{3} + 1) \\ c = \sqrt{3} \\ then \: d = \sqrt{ ({ \sqrt{3} + 1)}^{2} - 4 \times 1 \times \sqrt{3} } \\ = \sqrt{3 + 2 \sqrt{3} + 1 - 4 \sqrt{3} } \\ = \sqrt{3 - 2 \sqrt{3} + 1 } \\ = \sqrt{( \sqrt{3} - 1) ^{2} } \\ = \sqrt{3} - 1 \\ x1 = \frac{ - \sqrt{3} + 1 - \sqrt{3 } + 1}{2} \\ = - \sqrt{3} + 1 \\ and \\ x2 = (- \sqrt{3} + 1 + \sqrt{3} - 1) \div 2 = 0$