Question

Find a quadratic polynomial whose coefficient are rational and one zero is $5+\sqrt{7}$.

Answer 1

Solution:

Step 1: Find the second root:

Given: $5+\sqrt{7}$ is a zero of the quadratic polynomial

$\therefore (x-(5+\sqrt{7}\left)\right)$is a factor of quadratic polynomial

If $5+\sqrt{7}$ is a root, then $5-\sqrt{7}$ is also a root. $\left(\because \mathrm{conjugate}\mathrm{roots}\mathrm{occur}\mathrm{in}\mathrm{pairs}\right)$

$\therefore (x-(5-\sqrt{7}\left)\right)$ is also a factor of the polynomial

Step 2: Find the quadratic polynomial:

$$\begin{gathered} quadraticpolynomialisofform=k\left(\mathrm{x}-a\right)\left(\mathrm{x}-b\right) \\ =\left(\mathrm{x}-\left(5+\sqrt{7}\right)\right)\left(\mathrm{x}-\left(5-\sqrt{7}\right)\right) \\ ={\mathrm{x}}^2-(5+\sqrt{7}+5-\sqrt{7})\mathrm{x}+\left(5+\sqrt{7}\right)\left(5-\sqrt{7}\right) \\ ={\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{10}\mathbf{x}\mathbf{+}\mathbf{18} \end{gathered}$$

Final answer: Hence, the required polynomial is ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{10}\mathbf{x}\mathbf{+}\mathbf{18}\mathbf{.}$