Question

find the roots of the following quadratic equations if they exist with the method of completing the square
plz solve this question ​

Answer 1

Divide the equation by √2 we get.

x² - (3/√2) x - 2 = 0

$</p><p></p><p>\rightarrow \quad x^{2} - \frac{3}{\sqrt{2}}x - 2 = 0 </p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \left( x \right)^{2} - 2 \times \frac{3}{2\sqrt{2}} + \left( \frac{3}{2\sqrt{2}} \right)^{2} = 2 + \left( \frac{3}{2\sqrt{2}} \right)^{2} </p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \left( x - \frac{3}{2\sqrt{2}} \right)^{2} = 2 + \frac{9}{8} </p><p></p><p>\\ \\ </p><p></p><p>\rightarrow \quad \left( x - \frac{3}{2\sqrt{2}} \right)^{2} = \frac{16 + 9}{8}</p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \sqrt{\left( x - \frac{3}{2\sqrt{2}} \right)^{2} } = \sqrt{\frac{25}{8}}</p><p></p><p>\\ \\</p><p></p><p>\mathtt{Positive:} </p><p></p><p>\\ x_{1} = \frac{5}{2\sqrt{2}} + \frac{3}{2\sqrt{2}} </p><p>\\ \\</p><p>x_{1} = \frac{\cancel{8}}{\cancel{2}\sqrt{2}}</p><p></p><p>\\ \\</p><p></p><p>x_{1} = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} </p><p></p><p>\\ \\</p><p></p><p>x_{1} = 2\sqrt{2}</p><p></p><p></p><p>\\ \\</p><p></p><p>\mathtt{Negative:} </p><p></p><p>\\ </p><p></p><p>x_{2} = \frac{3}{2\sqrt{2}} - \frac{5}{2\sqrt{2}} </p><p></p><p>\\ \\</p><p></p><p>x_{2} = \frac{3-5}{2\sqrt{2}}</p><p></p><p>\\ \\</p><p></p><p>x_{2} = \frac{-\cancel{2}}{\cancel{2}\sqrt{2}}</p><p></p><p>\\ \\</p><p></p><p>x_{2} = - \frac{1}{\sqrt{2}}</p><p></p><p>$

$\bold{Zeroes \; are \; 2\sqrt{2} \; and \; - \frac{1}{\sqrt{2}} }$