Answer
385.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)