Question

find quardtic equation
whose one of zeroes is 2+ ✓3​

Answer 1

Step-by-step explanation:

If 2+sqrt(3) is a zero, so is the conjugate 2-sqrt(3) .

Also, if a is a zero, then (x-a) is a factor, thus the factors of the quadratic are:

(x-2+sqrt(3))(x-2-sqrt(3))

Multiplying we get:

x^2-2x-xsqrt(3)-2x+4+2sqrt(3)+xsqrt(3)-2sqrt(3)-3

Adding like terms:

x^2-4x+1

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The monic quadratic polynomial with rational coefficients is:

x^2-4x+1

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** Checking we can use the quadratic formula to find the roots(zeros):

x=(4+-sqrt(16-4(1)(1)))/(2(1))

=(4+-sqrt(12))/2

=(4+-2sqrt(3))/2

=2+-sqrt(3)

and we see that 2+sqrt(3) is a zero.

Answer 2

Answer:  $x^{2} -4x+1$

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Step-by-step explanation:

One root is $\alpha =2+\sqrt{3}$

second root will be $\beta =2-\sqrt{3}$

$\alpha +\beta =2+\sqrt{3} +2-\sqrt{3}\\\alpha +\beta =4$

$\alpha \beta =(2+\sqrt{3})(2-\sqrt{3} ) \\\alpha \beta =4-3\\\alpha \beta =1$

sum of roots = $\alpha +\beta = - \frac{b}{a}$

$\alpha +\beta =-\frac{b}{a} =4$   ----------> (1.)

product of roots = $\alpha \beta =\frac{c}{a}$

$\alpha \beta =1 =\frac{c}{a}$  ------------> (2.)

.: from (1.) and (2.) => a = 1, b = -4 , c = 1

so quadratic equation is of the form  =>   $ax^{2} +bx+c=0$

.:  $x^{2} -4x+1=0$