Question

find the zeroes of 2x square -3x+1 and verify the relationship between the zeroes and the coefficients ​

Answer 1

Answer:

${x}^{2} - 3x + 1 = 0 \: \: has \: zeros \: \alpha \: \: and \: \beta$

$\alpha + \beta = 3 \: \: \: \: and \: \: \: \alpha \beta = 1$

now,

$x = \frac{3 + \sqrt{9 - 4} }{2} \: \: or \: \: \frac{3 - \sqrt{9 - 4} }{2}$

$x = \frac{3 + \sqrt{5} }{2} \: \: or \: \: \frac{3 - \sqrt{5} }{2}$

$\frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2} = \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2} = 3 = \alpha + \beta$

$\frac{3 + \sqrt{5} }{2} \times \frac{3 - \sqrt{5} }{2} = \frac{9 - 5}{2 \times 2} = 1 = \alpha \beta$