Question

Solve by quadratic formula or completion of square

$17. \: x {}^{2} - (1 + \sqrt{2})x + \sqrt{2} = 0$
Ans: 1 , √2

Class:- 10

❎ Don't Spamming ❎

Answer 1

Hey!!!!

Good Evening

Difficulty Level : High

Chances of being asked in Board : 70%

________________

Refer to the attachment

_________________

Hope this helps ✌️

Answer 2

x² - (1 + √2)x + √2 = 0



$\Large{\sf{x^2 - \frac{2*X*(1 + \sqrt{2})}{2} + \frac{(1 + \sqrt{2})^2}{4} - \frac{(1 + \sqrt{2})^2}{4} + \sqrt{2} = 0}}$


$\Large{\sf{[x - (1 + \sqrt{2})/2]^2 = \frac{1 + 2 + 2\sqrt{2} - 4\sqrt{2}}{4} = 0}}$


$\Large{\sf {[ x - (1 + \sqrt{2})/2]^2 = \frac{(1 - \sqrt{2})^2}{4}}}$


$\Huge{\sf{taking \:\:Square\:\: Root \:\:both \:\:side}}$


$\Large{\sf{[x - (1 + \sqrt{2})/2] = ± \frac{1 - \sqrt{2}}{2}}}$



$\Huge{\sf{taking (-ve)}}$

$\Large{\sf{x = (1 + \sqrt{2})/2 - (1 - \sqrt{2})/2}}$

$\Large{\sf{x = (1 + \sqrt{2} - 1 + \sqrt{2})/2}}$

$\Large{\sf{x = 2\sqrt{2}/2 = \sqrt{2}}}$

$\Huge{\sf{taking (+ve)}}$

$\Large{\sf{x = (1 + \sqrt{2})/2 + (1 - \sqrt{2})/2}}$

$\Large{\sf{x = (1 + \sqrt{2} + 1 - \sqrt{2})/2}}$

$\Large{\sf{x = 2/2 = 1 }}$

$\Huge{\sf{Hence, \:\:x = 1 , \:\:x = \sqrt{2}}}$