Question

X square + X by root 2 + 1 is equal to zero​

Answer 1

Answer:

⇒ Factors : $(x + \frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}})$and $(x + \frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}})$

⇒ Zeroes : $x = -\frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}}$ (or) $x = -\frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}}$

Step-by-step explanation:

Given :

To find the factors & zeroes of the polynomial,

$x^2 + \frac{x}{\sqrt{2}} + 1 = 0$

Solution :

$x^2 + \frac{x}{\sqrt{2}} + 1 = 0$

⇒ $x^2 + (\frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}})x +(\frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}})x + 1 = 0$

⇒ $x(x + \frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}}) + (\frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}})(x +\frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}})=0$

⇒ $(x + \frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}})(x + \frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}})=0$

⇒ Factors : $(x + \frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}})$]and $(x + \frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}})$

⇒ Zeroes : $x = -\frac{1}{2\sqrt{2}}-\frac{i\sqrt{7}}{2\sqrt{2}}$ (or) $x = -\frac{1}{2\sqrt{2}}+\frac{i\sqrt{7}}{2\sqrt{2}}$