Question

solve for x:sqrt(2)x^2-3/sqrt(2)x+1/sqrt(2)

Answer 1

Quadratic formula = $\frac{-b \pm \sqrt{ b^{2}-4ac } }{2a}$

= $\frac{ \frac{3}{ \sqrt{2} }\pm \sqrt{ \frac{9}{2} } -4( \sqrt{2)}( \frac{1}{ \sqrt{2} }) }{2( \sqrt{2})}$

=$\frac{ \frac{3}{ \sqrt{2}} \pm \sqrt{ \frac{9}{2} -4 } }{2 \sqrt{2} }$

=$\frac{ \frac{3}{ \sqrt{2} }\pm \sqrt{ \frac{1}{2} } }{2 \sqrt{2} }$

=$\frac{ \frac{3 \pm 1}{ \sqrt{2} } }{2 \sqrt{2} }$

=$\frac{3\pm1}{4}$

x = $\frac{3+1}{4}$    x = $\frac{3-1}{4}$

x = 1, 1/2

Answer 2

To solve for x:
$\sqrt{2} x^{2} - \frac{3}{ \sqrt{2} }x+ \frac{1}{ \sqrt{2} }=0\\ \\ \Rightarrow \frac{1}{ \sqrt{2} }(2 x^{2} -3x+1)=0\\ \\ \Rightarrow2 x^{2} -3x+1= \sqrt{2} \times 0=0 \\ \\ \Rightarrow 2 x^{2} -2x-x+1=0\\ \\ \Rightarrow 2 x(x-1)-1(x-1)=0\\ \\ \Rightarrow (2x-1)(x-1)=0 \\ \\ \Rightarrow 2x-1=0\ \ or\ x-1=0\\ \\ \Rightarrow 2x=1\ \ or\ x=1\\ \\ \Rightarrow x= \frac{1}{2} \ \ or\ x=1$