
Consider the given expression \[4{{y}^{2}}+4y+1\ =\ 0\], then \[y\ =\ 0\].
(A). Yes
(B). No
(C). Ambiguous
(D). Data insufficient
Answer
597k+ views
- Hint: We can do it by factorizing with the middle term by factorizing method and from that, we will get the value of y then compare with the value of y given and write the correct option.
Complete step-by-step solution -
In the question we are given a quadratic equation \[4{{y}^{2}}+4y+1\ =\ 0\] and then we are given \[y\ =\ 0\] and we have to find \[y\ =\ 0\] satisfy or not.
If the quadratic equation is \[a{{y}^{2}}+by+c\ =\ 0\]
Then \[y\ =\ \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Where a is coefficient of \[{{y}^{2}}\], b is coefficient of \[y\] and c is constant.
In the equation \[4{{y}^{2}}+4y+1\ =\ 0\] can be considered as quadratic equation which can be rearranged in standard form as \[a{{y}^{2}}+by+c\ =\ 0\].
Here $y$ represents unknown and $a,b,c$ are known numbers where $a\ne 0$, otherwise it becomes a linear equation.
Consider the given expression \[4{{y}^{2}}+4y+1\ =\ 0\], this can be rewritten as
\[4{{y}^{2}}+2y+2y+1\ =\ 0\]
On factorizing we get,
\[\begin{align}
& \Rightarrow 2y\left( 2y+1 \right)+1\left( 2y+1 \right)=0 \\
& \left( 2y+1 \right)\left( 2y+1 \right)\ =\ 0 \\
\end{align}\]
So, the value of \[y\ =\ -\dfrac{1}{2}\], but the value of y given as 0 so, the answer is NO.
So, the correct option is ‘B’.
Note: Another approach is - we can also get the answer by just putting the value of \[y\ =\ 0\] in equation \[4{{y}^{2}}+4y+1\ =\ 0\] to find the answer whether it satisfies or not.
Complete step-by-step solution -
In the question we are given a quadratic equation \[4{{y}^{2}}+4y+1\ =\ 0\] and then we are given \[y\ =\ 0\] and we have to find \[y\ =\ 0\] satisfy or not.
If the quadratic equation is \[a{{y}^{2}}+by+c\ =\ 0\]
Then \[y\ =\ \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Where a is coefficient of \[{{y}^{2}}\], b is coefficient of \[y\] and c is constant.
In the equation \[4{{y}^{2}}+4y+1\ =\ 0\] can be considered as quadratic equation which can be rearranged in standard form as \[a{{y}^{2}}+by+c\ =\ 0\].
Here $y$ represents unknown and $a,b,c$ are known numbers where $a\ne 0$, otherwise it becomes a linear equation.
Consider the given expression \[4{{y}^{2}}+4y+1\ =\ 0\], this can be rewritten as
\[4{{y}^{2}}+2y+2y+1\ =\ 0\]
On factorizing we get,
\[\begin{align}
& \Rightarrow 2y\left( 2y+1 \right)+1\left( 2y+1 \right)=0 \\
& \left( 2y+1 \right)\left( 2y+1 \right)\ =\ 0 \\
\end{align}\]
So, the value of \[y\ =\ -\dfrac{1}{2}\], but the value of y given as 0 so, the answer is NO.
So, the correct option is ‘B’.
Note: Another approach is - we can also get the answer by just putting the value of \[y\ =\ 0\] in equation \[4{{y}^{2}}+4y+1\ =\ 0\] to find the answer whether it satisfies or not.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Who gave "Inqilab Zindabad" slogan?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

