Write the quadratic equation that has roots (√3 +1)/2 and (√3 −1)/2 , if its coefficient with x^2 is equal to 1.

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bayosanuade bayosanuade · Answer 1 · 2020-04-18T18:19:15+02:00

Answer: [tex]x^{2} - \sqrt{3}x + \frac{1}{2} = 0[/tex]

Step-by-step explanation:

The formula for finding quadratic equation whose roots are given is :

[tex]x^{2} - sum of roots(x) + product = 0[/tex]

[tex]sum = \frac{\sqrt{3}+1}{2} +\frac{\sqrt{3}-1}{2}[/tex]

[tex]= \frac{\sqrt{3}+1+\sqrt{3}-1}{2}[/tex]

[tex]= \frac{2\sqrt{3}}{2}[/tex]

[tex]= \sqrt{3}[/tex]

[tex]product = \frac{\sqrt{3}+1}{2}[/tex] x [tex]\frac{\sqrt{3}-1}{2}[/tex]

[tex]= \frac{2}{4}[/tex]

[tex]= \frac{1}{2}[/tex]

substituting into the formula, we have:

[tex]x^{2} - \sqrt{3}x + \frac{1}{2} = 0[/tex]

ankitprmr2 ankitprmr2 · Answer 2 · 2021-10-11T13:51:37+02:00

Answer:

The required quadratic equation is

[tex]x^2-\sqrt{3} x+0.5 = 0[/tex]

Step-by-step explanation:

We know that the quadratic equation in terms of sum of roots and product of roots is given by,

[tex]x^2-(\alpha +\beta )x+(\alpha \beta )=0[/tex] ------- (1)

Given :

[tex]\alpha = \dfrac{\sqrt{3}+1 }{2}[/tex]

[tex]\beta = \dfrac{\sqrt{3} -1}{2}[/tex]

Calculation :

sum of roots -

[tex]\alpha +\beta =(\dfrac{\sqrt{3}+1 }{2})+(\dfrac{\sqrt{3}-1 }{2}) = \sqrt{3}[/tex] ------ (2)

product of roots -

[tex]\alpha \beta =(\dfrac{\sqrt{3} +1}{2})(\dfrac{\sqrt{3}-1 }{2}) = \dfrac{1}{2}=0.5[/tex] ------ (3)

from equation (1), (2) and (3) we get our required quadratic equation, which is

[tex]x^2-\sqrt{3} x+0.5 = 0[/tex]

For more information, refer the link given below

https://brainly.com/question/1162060?referrer=searchResults