Question

Solve for x; x²-(√2+1)x+√2 = 0

Answer 1

Answer:

hey bro , well couldn't have started it off by saying , well this is in the nick of time..

x2-rootx-x+root=0(upon using distributive property )

rearrange it now , x2-x-root2x+root2-0

x(x-1)-root2(x-1)

therefore , (x-1)(x-root2)=0

x=root2

x=1

there you go buddy.

Step-by-step explanation:

Answer 2

By Sridharachariya formula, we get,
$x = \frac{ - ( - ( \sqrt{2} - 1 )) + - \sqrt{ {( - ( \sqrt{2} - 1 ))}^{2} - 4 \times 1 \times \sqrt{2} }}{2 \times 1} \\ \: \: \: = \frac{ \sqrt{2} - 1 + - \sqrt{ ({ \sqrt{2} - 1)}^{2} - 4 \sqrt{2} } }{2} \\ \: \: \: = \frac{ \sqrt{2} - 1 + - \sqrt{2 - 2\sqrt{2} + 1 - 4 \sqrt{2} } }{2} \\ \: \: \: = \frac{ \sqrt{2} - 1 + - \sqrt{3 - 6 \sqrt{2} } }{2}$
The square root of a negetive number does not exists in the set of Real Numbers.
So, x does not exist in real number. Therefore, this equation has no solutions.