# If \[a{{x}^{2}}+bx+c=0\] is satisfied by every values of x, then

A. b,c = 0

B. c= 0

C. a = 0

D. a=b=c=0

Last updated date: 20th Mar 2023

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Answer

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Hint: If any polynomial equation of any degree satisfied for every value of it’s variable then it would become an identity and all it’s coefficients must equal to 0.

Complete step-by-step answer:

In the above question \[a{{x}^{2}}+bx+c=0\] is satisfied by every value of x.

Now, we will check all the option one by one which is shown below;

A. b,c = 0

So, the above equation will be as follow;

$\begin{align}

& \Rightarrow a{{x}^{2}}=0 \\

& \Rightarrow x=0,0 \\

\end{align}$

Here, only one value is there which satisfies the equation.

So, this option is false.

Again, we have option B. c=0

Now, equation will be as follow;

\[\begin{align}

& \Rightarrow a{{x}^{2}}+bx=0 \\

& \Rightarrow x(ax+b)=0 \\

& \Rightarrow x=0,x=\dfrac{-b}{a} \\

\end{align}\]

Here, there are only two values that satisfy the equation.

So, this option is false.

Again, we have option C. a=0 .

Now, the above equation will be as follow;

\[\begin{align}

& \Rightarrow bx+c=0 \\

& \Rightarrow x=\dfrac{-c}{b} \\

\end{align}\]

Here, also there is only one value that satisfies the equation.

So, this option is also false.

Again, we have option D. a=b=c=0.

Now, the equation will be as follow;

\[\begin{align}

& \Rightarrow 0\times {{x}^{2}}+0\times x+0=0 \\

& \Rightarrow 0=0 \\

\end{align}\]

Here, all the values of x will satisfy the equation.

Hence, the correct answer will be a=b=c=0.

Therefore, the correct option of the above question will be option D.

NOTE:

Just remember the point that if a polynomial equation satisfies every value for it’s variable then it would become an identity.

Complete step-by-step answer:

In the above question \[a{{x}^{2}}+bx+c=0\] is satisfied by every value of x.

Now, we will check all the option one by one which is shown below;

A. b,c = 0

So, the above equation will be as follow;

$\begin{align}

& \Rightarrow a{{x}^{2}}=0 \\

& \Rightarrow x=0,0 \\

\end{align}$

Here, only one value is there which satisfies the equation.

So, this option is false.

Again, we have option B. c=0

Now, equation will be as follow;

\[\begin{align}

& \Rightarrow a{{x}^{2}}+bx=0 \\

& \Rightarrow x(ax+b)=0 \\

& \Rightarrow x=0,x=\dfrac{-b}{a} \\

\end{align}\]

Here, there are only two values that satisfy the equation.

So, this option is false.

Again, we have option C. a=0 .

Now, the above equation will be as follow;

\[\begin{align}

& \Rightarrow bx+c=0 \\

& \Rightarrow x=\dfrac{-c}{b} \\

\end{align}\]

Here, also there is only one value that satisfies the equation.

So, this option is also false.

Again, we have option D. a=b=c=0.

Now, the equation will be as follow;

\[\begin{align}

& \Rightarrow 0\times {{x}^{2}}+0\times x+0=0 \\

& \Rightarrow 0=0 \\

\end{align}\]

Here, all the values of x will satisfy the equation.

Hence, the correct answer will be a=b=c=0.

Therefore, the correct option of the above question will be option D.

NOTE:

Just remember the point that if a polynomial equation satisfies every value for it’s variable then it would become an identity.

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