Question

find a quadratic polynomial,the sum and product of whose zeroes are 6 and 6 respectively.Hence find the zeroes

Answer 1

$\alpha + \beta = 6 \\ \alpha \beta = 6 \\Polynomial = {x}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} - 6x + 6 \\ \\ {x}^{2} - 6x + 6 = 0$

$D = {b}^{2} - 4ac \\ \: \: \: \: \: = {6}^{2} - 4 \times 1 \times 6 \\ \: \: \: \: \: = 36 - 24 \\ \: \: \: \: \: = 12$

$\sqrt{D} = \sqrt{12} = 2 \sqrt{3}$

$x = \frac{ - b + \sqrt{D} }{2a} \: \: or \: \: \frac{ - b - \sqrt{D} }{2a} \\ \: \: \: = \frac{ 6 + 2 \sqrt{3} }{2} \: \: or \: \: \frac{6 - 2 \sqrt{3} }{2} \\ \: \: \: = 3 + \sqrt{3 } \: \: \: or \: \: \: 3 - \sqrt{3}$

$So, \: the \: zeroes \: are \: 3 + \sqrt{3} \: and \: 3 - \sqrt{3} .$