# Value of $c$ in $x^2-\sqrt2x+c=0$

If the roots $$\alpha$$ and $$\beta$$ of the equation, $$x^2-\sqrt2x+c=0$$ are complex for some real numbers $$c\ne 1$$ and $$|\frac{\alpha-\beta}{1-\alpha\beta}|=1$$ then a value of $$c$$ is

Squaring both sides, I get $$(\alpha+\beta)^2-4\alpha\beta=(1-\alpha\beta)^2$$

Putting $$\alpha+\beta=\sqrt2$$ and $$\alpha\beta=c$$, I get $$c^2+2c-1=0\implies c=-1\pm\sqrt2$$.

Though, the answer is given as $$3-\sqrt6$$.

Also, I wonder if the roots being complex have any bearing on the solution.

• $|x|^2=x^2$ does not necessarily hold. May 21 '20 at 7:27
• @mathlove oh thanks. We need to use $|z|^2=z\overline z$. That is why they have mentioned the roots to be complex numbers! May 21 '20 at 7:33
• Since the roots are given complex, they will be conjugates of each other May 21 '20 at 7:33
• @DhanviSreenivasan I wonder how to use that info to my advantage here? May 21 '20 at 7:36
• Go ahead, that should help people who may have similar problems in the future May 21 '20 at 7:45

With the help of the comments above, I am able to solve the question. Here is my solution:

Since the roots are complex, so when we square the given expression, we need to use $$|z|^2=z\overline z$$. Thus, $$\frac{\alpha-\beta}{1-\alpha\beta}\cdot\frac{\overline\alpha-\overline\beta}{1-\overline\alpha\overline\beta}=1$$

And since, complex roots are conjugate of each other. So, by replacing $$\overline\alpha$$ with $$\beta$$ and $$\overline\beta$$ with $$\alpha$$, we get $$(\frac{\alpha-\beta}{1-\alpha\beta})^2=-1$$

Thus, quadratic in $$c$$ becomes $$c^2-6c+3=0$$

Thus, $$c=3\pm\sqrt6$$

Notice, you can't take squares of $$\left|\frac{\alpha-\beta}{1-\alpha\beta}\right|=1$$ in this case as $$(\alpha-\beta)$$ is pure imaginary

The roots $$\alpha$$ & $$\beta$$ of quadratic equation: $$x^2-\sqrt2x+c=0$$ will be complex only if $$B^2-4AC<0$$ i.e. $$(-\sqrt2)^2-4(1)(c)<0\iff c>\frac12$$ $$\alpha+\beta=\frac{-(-\sqrt2)}{1}=\sqrt2 \quad \text{&}\quad \alpha\beta=c$$ It's worth noticing that the difference of complex roots $$\alpha-\beta$$ will be pure imaginary as they are conjugate $$\therefore \alpha-\beta=\pm\sqrt{(\alpha+\beta)^2-4\alpha\beta}=\pm\sqrt{(\sqrt2)^2-4c}=\pm\sqrt{2-4c}=\pm i\sqrt{4c-2}$$ Setting corresponding values in $$\left|\frac{\alpha-\beta}{1-\alpha\beta}\right|=1$$, $$\left|\frac{\pm i\sqrt{4c-2}}{1-c}\right|=1$$ $$\sqrt{4c-2}=|1-c|$$ Taking squares as both the sides are positive, $$(\sqrt{4c-2})^2=(1-c)^2$$ $$c^2-6c+3=0$$ $$c=3\pm\sqrt6$$ Both the above real values of $$c$$ duly satisfy, $$c\in \left(\frac12,1\right)\cup \left(1,\infty\right)$$

• "taking squares of an equation is valid iff both the sides are positive". I think it would work otherwise too. If $x=-2$, $x^2=4$. May 21 '20 at 9:18
• Yes, you are right. But when both sides have unknown variables then it wouldn't necessarily hold. May 21 '20 at 9:24