Question

If one zero of polynomial x2-4x+1 is 2+root 3 wdh other zero

Answer 1

Answer:

The other zero is P=2-√3

Step-by-step explanation:

$x^{2}-4 x+1$ is the given equation,

One of the root is 2+√3  and  another root is 0 .

Let p(x)=  $x^{2}-4 x+1$

Compare p(x)  with  $a x^{2}+b x+c$ (linear equation)

a=1, b=-4 & c=1

One zero = 2+√3

Let second zero = p

Sum of zeros = -b/a      

Substitute in the form of  $a x^{2}+b x+c$ P+(2+√3)=-(-4)                                                                  

P=4-(2+√3)                                                                          

P=2-√3          

Product of zero = c/a

(2+√3)p=1

p=1/(2√3)

On rationalizing

P=2-√3/[(2+√3)(2-√3)]

P=2-√3/[22-〖√3〗^2]

P= 2-√3/4-3

P=2-√3

Therefore second zero of p(x)

P=2-√3

Answer 2

Answer:

$Other \:zero = 2-\sqrt{3}$

Step-by-step explanation:

$Let\:\alpha \:and\:\beta\:are \\two \: zeroes \:of\: quadratic \: polynomial \\p(x)=x^{2}-4x+1$

$\alpha = 2+\sqrt{3}\:(given)$

$Now,\\compare\:p(x)\:with \:ax^{2}+bx+c \:we\:get$

$a=1, \:b=-4,\: c = 1$

$Sum\:of\: the\:zeroes =\frac{-b}{a}$

$\implies \alpha+\beta=\frac{-(-4)}{1}$

$\implies 2+\sqrt{3}+\beta = 4$

$\implies \beta = 4-2-\sqrt{3}$

$\implies \beta=2-\sqrt{3}$

Therefore,

$Other \:zero = 2-\sqrt{3}$

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