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# If $a{{x}^{2}}+bx+c=0$ is satisfied by every values of x, thenA. b,c = 0B. c= 0C. a = 0D. a=b=c=0

Last updated date: 20th Mar 2023
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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.

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.