Question

The quadratic equation having rational coefficients whose one root is 3+root2

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{One root of a quadratic equation is}\;\mathsf{3+\sqrt2}$

$\underline{\textbf{To find:}}$

$\textsf{The quadratic equation having rational coefficients whose}$

$\mathsf{one\;root\;is\;3+\sqrt2}$

$\underline{\textbf{Solution:}}$

$\mathsf{Take,\;\;x=3+\sqrt2}$

$\textsf{This can be written as,}$

$\mathsf{x-3=\sqrt2}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(x-3)^2=2}$

$\mathsf{x^2+9-6x=2}$

$\implies\boxed{\bf\,x^2-6x+7=0}$

$\underline{\textbf{Answer:}}$

$\textbf{The required quadratic equation is}\;\bf\;x^2-6x+7=0$

Answer 2

Answer:

The required quadratic equation is

${x}^{2} - 6x + 5 = 0$

Step-by-step explanation:

Given that,

There is an quadratic equation having rational coefficients with one root as

$\alpha = 3 + \sqrt{2}$

According to theory of equations,irrationalroots of a quadratic equation exist in form of conjugate pairs.Therefore,the other root of the equation is

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

Now the sum of roots is

$\alpha + \beta = 3 + \sqrt{2} + 3 - \sqrt{2} = 6$

Product of roots is

$\alpha \beta = (3 + \sqrt{2} )(3 - \sqrt{2}) = 5$

Now the quadratic equation with the above two roots is

${x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ = > x {}^{2} - 6x + 5 = 0$

#SPJ3