Question

If \[f\left( 0 \right) = 1\] , \[f\left( 1 \right) = 5\] and \[f\left( 2 \right) = 11\] ,then the equation of polynomial of degree two is
\[\left( 1 \right){\text{ }}{x^2} + 1 = 0\]
\[\left( 2 \right){\text{ }}{x^2} + 3x + 1 = 0\]
\[\left( 3 \right){\text{ }}{x^2} - 2x + 1 = 0\]
\[\left( 4 \right)\] None of these

Answer 1

Hint: The general form of the equation of polynomial of degree two is \[a{x^2} + bx + c = 0\] or \[f\left( x \right) = a{x^2} + bx + c\] .Therefore one by one apply the given conditions in \[f\left( x \right) = a{x^2} + bx + c\] by which you will be able to get the values of a, b and c. Then substitute these values in the equation \[a{x^2} + bx + c = 0\] to get the desired result.

Complete step-by-step answer:
Quadratic equations are the polynomial equations of degree two in one variable of type \[f\left( x \right) = a{x^2} + bx + c\] where a, b, c are real numbers and \[a \ne 0\] .It is the general form of a quadratic equation where ‘a’ is called the leading coefficient and c is called the absolute term of f(x).
Therefore, let the function f(x) is given by
\[f\left( x \right) = a{x^2} + bx + c\] -------- (i)
It is given to us that the value of function at \[x = 0\] is one that is \[f\left( 0 \right) = 1\] .By putting x equal to zero in equation (i) we get
\[ \Rightarrow a{\left( 0 \right)^2} + b\left( 0 \right) + c = 1\]
\[ \Rightarrow c = 1\] --------- (ii)
By putting the value of equation (ii) in equation (i) we get
\[f\left( x \right) = a{x^2} + bx + 1\] -------- (iii)
It is also given to us that \[f\left( 1 \right) = 5\] .So by putting x equal to one in equation (iii) we get
\[ \Rightarrow a{\left( 1 \right)^2} + b\left( 1 \right) + 1 = 5\]
On further solving we get
\[ \Rightarrow a + b = 5 - 1\]
\[ \Rightarrow a + b = 4\] -------- (iv)
In the question it is also given that \[f\left( 2 \right) = 11\] .So by putting x equal to two in equation (iii) we get
\[ \Rightarrow a{\left( 2 \right)^2} + b\left( 2 \right) + 1 = 11\]
By opening the brackets we get
\[ \Rightarrow 4a + 2b + 1 = 11\]
On further solving we get
\[ \Rightarrow 4a + 2b = 11 - 1\]
\[ \Rightarrow 4a + 2b = 10\] ------- (v)
On solving equations (iv) and (v) we get
\[ \Rightarrow 2b = 6\]
That is
\[ \Rightarrow b = \dfrac{6}{2}\]
\[ \Rightarrow b = 3\] ----------- (vi)
By putting this value of b in equation (iv) we get
\[ \Rightarrow a + 3 = 4\]
On shifting three to the right hand side we get
\[ \Rightarrow a = 4 - 3\]
\[ \Rightarrow a = 1\] -------- (vii)
Thus by putting the values from equations (vi) and (vii) in the equation (iii) we get
\[f\left( x \right) = {x^2} + 3x + 1\]
Hence the correct option is \[\left( 2 \right){\text{ }}{x^2} + 3x + 1 = 0\]
So, the correct answer is “Option 2”.

Note: Remember that the general form of degree two equations is \[a{x^2} + bx + c = 0\] where a, b, c are the real numbers. Keep in mind that quadratic equations and degree two equations are the same things because a quadratic polynomial is a type of polynomial which has a degree of two. So, a quadratic polynomial has a degree of two.