Question

Find a quadratic polynomial whose zeroes are 5 + root 2 and 5 minus root 2

Answer 1

Answer:

x² - 10x + 23

Step-by-step explanation:

given that,

zeros of the quadratic equation,

5 + √2 and 5 - √2

we know that,

quadratic polinomial when zeros are given then,

equation = x² - sum of zeros × x + product of zeros

here,

sum of zeros = 5 + √2 + 5 - √2

= 10

and

product of zeros = (5 + √2)(5 - √2)

= 23

now,

putting the values,

x² - 10x + 23

so,

the required quadratic equation will be,

x² - 10x + 23

Answer 2

Answer:

$\bigstar\: \underline{\boxed{\mathsf{x^2 -10x + 23 }}}$ $\bigstar$

Step-by-step explanation:

$\underline{\underline{\textsf{Given That...}}}$

$\mathsf{zeroes \: or \: root \: of \: a \: quadratic \: equation \: are \: 5 + \sqrt{2} \: and \: 5 - \sqrt{2} }$

$\underline{\underline{\mathfrak{Solution}}}$

$\textsf{Now we know that General form of quadratic polynomial is }$

$\bigstar \boxed{\mathsf{ax^2 + bx + c }}$

$\mathsf{Where\: 'a' , 'b' \:and \:'c'\: are\: coefficient\: of\: x^2 , x \:and \:constant \: respectively.}$

$\textsf{Now we know that,}$

$\bigstar \mathsf{Sum\: of\:zeroes\:=\: \dfrac{-b}{a}}$

$\textsf{and}$

$\bigstar \mathsf{Product\: of\:zeroes\:=\: \dfrac{c}{a}}$

$\underline{\textsf{Therefore here we have..}}$

$\implies \mathsf{(5+\sqrt{2})(5-\sqrt{2}) \:=\:\dfrac{c}{a}}$

$\implies \mathsf{{(5)}^{2}-{(\sqrt{2}}^{2}\:=\:\dfrac{c}{a}}$ (By (a+b)(a-b)=a^2-b^2) )

$\implies \mathsf{25 - 2 \:=\:\dfrac{c}{a}}$

$\implies \mathsf{\dfrac{23}{1} \:=\:\dfrac{c}{a}}$ $\rule{40}{1}$ eq(1)

$\textsf{Now, we have,}$

$\mathsf{Sum\: of\:zeroes\:=\: \dfrac{-b}{a}}$

$\textsf{That is}$

$\implies \mathsf{5 +\sqrt{2} + 5-\sqrt{2} = \dfrac{-b}{a}}$

$\implies \mathsf{5 + 5\: =\: \dfrac{-b}{a}}$

$\implies \mathsf{\dfrac{10}{1} = \dfrac{-b}{a} }$ $\rule{40}{1}$ eq(2)

$\textsf{Now by comparing eq (1) and eq (2) we have...}$

$\mathsf{ a = 1 , b= (-10)\: and \: c = 23 }$

$\textsf{Now putting the values in General form of quadratic polynomial we have,}$

$\implies \mathsf{1x^2 -10x + 23 }$

$\underline{\underline{\textsf{Therefore Required Quadratic polynomial is }}}$

$\underline{\boxed{\boxed{\mathsf{x^2 -10x + 23 }}}}$