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How do you solve this by completing the square for $2{{x}^{2}}+2x=0$?

seo-qna
Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: We have been given a quadratic equation of $x$ as $2{{x}^{2}}+2x=0$. We first try to form the square form of the given equation and find its root value from the square. We also use the quadratic formula to solve the value of the x. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete step-by-step solution:
We have been given the equation $2{{x}^{2}}+2x=0$. We need to form the square part in $2{{x}^{2}}+2x$.
The square form of subtraction of two numbers be ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
We have \[2{{x}^{2}}+2x={{\left( x\sqrt{2} \right)}^{2}}+2\times x\sqrt{2}\times \dfrac{1}{\sqrt{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\].
Forming the square, we get \[2{{x}^{2}}+2x={{\left( x\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\].
We get \[2{{x}^{2}}+2x={{\left( x\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=0\]. Taking solution, we get
\[\begin{align}
  & {{\left( x\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)}^{2}}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}} \\
 & \Rightarrow \left( x\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)=\pm \dfrac{1}{\sqrt{2}} \\
 & \Rightarrow x=0,-1 \\
\end{align}\].
Thus, the solution of the equation $2{{x}^{2}}+2x=0$ is \[x=0,-1\].
We find the value of $x$ for which the function $f\left( x \right)=2{{x}^{2}}+2x$. We can see $f\left( 0 \right)=2\times 0+2\times 0=0$. So, the root of the $f\left( x \right)=2{{x}^{2}}+2x=0$ will be the 0. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We can also do the same process for \[\left( x+1 \right)\].

Note: We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method.
In the given equation we have $2{{x}^{2}}+2x=0$. The values of a, b, c are $2,2,0$ respectively.
We put the values and get x as \[x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 0\times 2}}{2\times 2}=\dfrac{-2\pm \sqrt{4}}{4}=\dfrac{-2\pm 2}{4}=0,-1\].
The roots of the equation are rational numbers.
The discriminant value being square, we get the rational numbers a root value.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square. ${{b}^{2}}-4ac={{2}^{2}}-4\times 0\times 2=4$.
This is a square value. That’s why the roots are rational.