Re-write a quadratic equation in another form?

Question

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

I need to find the real solutions for this equation and write it in this form:

$$\frac{-\sqrt{A} \pm B}{C}$$

So when I work the problem out with the quadratic equation I get: $x = 0.118121$

I had no idea how to even put that in the form described so I assumed they didn't want me to solve but just put it in that form so I put it as follows:

$A = x^2$

$B = \sqrt{2}x$

$C = \frac{1}{2}$

That was wrong also. I'm not really sure what they're asking me to do. I know the answer is: $x = 0.118121$ but is it even possible to put it in the form described? Any help would be appreciated.

Answer 1

Form of any quadratic Equation is $ax^2+bx+c=0$

Here for you $ a=1 , b= \sqrt{2}$ and $c = -\frac{1}{2}$, $$x^2 + \sqrt{2}x + \left(-\frac{1}{2} \right)=0$$

Solution is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-\sqrt{2}\pm \sqrt{(\sqrt{2})^2-4\cdot 1 \cdot \left(-\frac{1}{2}\right)}}{2 \cdot 1}$$

$$x=\frac{-\sqrt{2}\pm \sqrt{2+2}}{2 }$$

$$x=\frac{-\sqrt{2}\pm \sqrt{4}}{2 }$$

$$x=\frac{-\sqrt{2}\pm 2}{2 }$$

So $$A=2 ,B =2 ,C=2$$

Answer 2

Hint: $x^2+\sqrt{2}x=\frac{1}{2}\implies2x^2+2\sqrt{2}x-1=0$ Now use the general way to solve an quadratic.