Question

Find the roots of the following quadratic (if they exist) by the method of completing the square.
x²-(√2+1)x+√2=0

Answer 1

The roots of the quadratic equation are $\sqrt{2}$ and 1

Explanation:

Given that the equation is $x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$

We need to determine the roots of the quadratic equation by the method of completing the square.

Let us take the term $\sqrt{2}$ to the RHS, we get,

$x^{2}-(\sqrt{2}+1) x=-\sqrt{2}$

Completing the square, we have,

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

Let us simplify the RHS term $\left(\frac{\sqrt{2}+1}{2}\right)^{2}-\sqrt{2}$

Thus, we have,

$\left(\frac{\sqrt{2}+1}{2}\right)^{2}-\sqrt{2}=\left(\frac{(\sqrt{2})^2+2(\sqrt{2})+ 1^2}{2^2}\right)-\sqrt{2}$

                       $=\left(\frac{2+2\sqrt{2}+1}{4}\right)-\sqrt{2}$

                       $=\left\frac{3+2\sqrt{2}}{4}\right-\sqrt{2}$

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

Substituting the value of RHS in equation (1) , we have,

$x^{2}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}-2\left(\frac{\sqrt{2}+1}{2}\right) x=\frac{3-2 \sqrt{2}}{4}$

Simplifying, we get,

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

Taking square root on both sides of the equation, we get,]

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

            $x=\frac{\sqrt{2}+1}{2} \pm \frac{\sqrt{2}-1}{2}$

Thus, the values of x are

$x=\frac{\sqrt{2}+1}{2} + \frac{\sqrt{2}-1}{2}$  and $x=\frac{\sqrt{2}+1}{2}- \frac{\sqrt{2}-1}{2}$

Let us simplify the term $x=\frac{\sqrt{2}+1}{2} + \frac{\sqrt{2}-1}{2}$ , we have,

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

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

$x=\sqrt{2}$

Also, simplifying the term $x=\frac{\sqrt{2}+1}{2}- \frac{\sqrt{2}-1}{2}$ , we get,

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

$x=\frac{2}{2}$

$x=1$

Thus, the roots of the equation are $\sqrt{2}$ and 1

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