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# Find the roots of the following quadratic equation (if they exist) by the method of completing the square. ${{x}^{2}}-\left( \sqrt{2}+1 \right)x+\sqrt{2}=0$(a) Exists, $\sqrt{2},1$(b) Exists, $-\sqrt{2},1$(c) Exists, $-\sqrt{2},-1$(d) None of these

Last updated date: 16th Jul 2024
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Hint: To find the roots of the given equation by completing the square method, simplify the terms on the left hand side of the given equation in the form of a complete square. Equate the left hand side of the equation to zero and simplify to find the roots.

We have to find the roots of the quadratic equation ${{x}^{2}}-\left( \sqrt{2}+1 \right)x+\sqrt{2}=0$ by method of completing the square. To do so, we will rearrange the terms on left hand side of the equation to make a perfect square and equate it to zero to find the roots.
We can rewrite the equation ${{x}^{2}}-\left( \sqrt{2}+1 \right)x+\sqrt{2}=0$ as ${{x}^{2}}-2\times \dfrac{1}{2}\left( \sqrt{2}+1 \right)x+\sqrt{2}=0$.
$\Rightarrow$ ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+\sqrt{2}=0$.
To complete the square, we will add the given equation by the square of $\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)$ on both sides.
$\Rightarrow$ ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+\sqrt{2}=0$ as ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+\sqrt{2}+{{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)}^{2}}={{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)}^{2}}$.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\Rightarrow$ ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+{{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)}^{2}}=\dfrac{1}{2}+\dfrac{1}{4}+2\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)-\sqrt{2}$.
Simplifying the right side of this equation, we have
$\Rightarrow$ ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+{{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\right)}^{2}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{\sqrt{2}}-\sqrt{2}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1-2}{\sqrt{2}}=\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{\sqrt{2}}$.

We also know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
So, we can write the left side of equation as ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+{{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)}^{2}}={{\left( x-\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right) \right)}^{2}}$.
Similarly, we can write the right side of equation as $\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{\sqrt{2}}={{\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)}^{2}}$.
Thus, we can write the equation as
$\Rightarrow$ ${{x}^{2}}-2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)x+{{\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)}^{2}}=\dfrac{1}{2}+\dfrac{1}{4}+2\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)-\sqrt{2}$ as ${{\left( x-\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right) \right)}^{2}}={{\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)}^{2}}$.
Taking square root on both sides, we have $x-\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2} \right)=\pm \dfrac{1}{\sqrt{2}}-\dfrac{1}{2}$.
Further simplifying the above equation, we have
$\Rightarrow$ $x=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}$
$\Rightarrow$ $x=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}-\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)$.
So, we have
$\Rightarrow$$x=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$
$\Rightarrow$ $x=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}-\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)=1$

Hence, the roots of the quadratic equation ${{x}^{2}}-\left( \sqrt{2}+1 \right)x+\sqrt{2}=0$ are $x=\sqrt{2},1$, which is option (a).

Note: There are multiple ways to solve this quadratic equation. We can also solve it using the factorization method by splitting the middle terms. We solved it by completing the square as it was given in the question. We can also check that the roots calculated by us are correct or not by substituting the roots in the quadratic equation and checking whether they satisfy the given equation or not.