Question

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

Answer 1

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.