Question

find the quadratic polynomial whose zeros are 2-√3 and -√3/2​

Answer 1

Answer:

Polynomial = x²-2+√3 - √3/2

Step-by-step explanation:

sum of zeros 2-√3

product of zeros = -√3/2

polynomial = x²- ( sum of zeros )x + (product of

zeros)

Polynomial = x²-2+√3 - √3/2

JAI SHREE KRISHNA

Answer 2

Step-by-step explanation:

$\alpha = 2 - \sqrt{3} \\ \beta = \frac{ - \sqrt{3} }{2} \\ sum \: of \: zeros \: ( \alpha + \beta )= 2 - \sqrt{3} + ( - \frac{ \sqrt{3} }{2} ) \\ \: \: \: = \frac{2 - \sqrt{3} }{1} - \frac{ \sqrt{3} }{2} \\ \: \: \: = \frac{4 - 2 \sqrt{3} - \sqrt{3} }{2} \\ = \frac{4 - 3 \sqrt{3} }{2} \\ \\ \\ product \: of \: zeros \: \alpha \beta = (2 - \sqrt{3} ) \times ( - \frac{ \sqrt{3} }{2} ) \\ = \frac{ - 2 \sqrt{3} + 3 }{2} \\ = \frac{3 - 2 \sqrt{3} }{2} \\ \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ = > {x}^{2} - \frac{4 - 3 \sqrt{3} }{2} x \: + \frac{3 - 2 \sqrt{2} }{2} = 0 \: ans$