Question

Write an example for a quadratic polynomial that has no real zeros.

Answer 1

Hint: In this question, we first need to write the general form of a quadratic equation given by \[a{{x}^{2}}+bx+c=0\]. Then find the discriminant of that quadratic equation which is given by the formula \[\sqrt{{{b}^{2}}-4ac}\]. Now, for the zeros of that quadratic polynomial to be not real the discriminant should be less than zero. Then take a, b, c values accordingly which satisfies the condition.

Complete step by step solution:
Quadratic Equation:
A polynomial of second degree is called a quadratic polynomial.
A quadratic polynomial \[f\left( x \right)\]when equated to zero is called quadratic equation.
i.e. \[a{{x}^{2}}+bx+c=0,a\ne 0\]
The values of the variable x which satisfy the quadratic equation are called zeros of the quadratic equation.
Direct Formula:
Quadratic equation \[a{{x}^{2}}+bx+c=0,a\ne 0\]has two roots, given by
\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Where, \[D=\sqrt{{{b}^{2}}-4ac}\]is called discriminant of the equation.
Nature of Roots:
Let the quadratic equation be \[a{{x}^{2}}+bx+c=0\], whose discriminant is D.
For \[a{{x}^{2}}+bx+c=0;a,b,c\in R\text{ and }a\ne 0\], if
(a) \[D<0\] then the zeros of the equation are complex
(b) \[D>0\] then the zeros of the equation are real and distinct
(c) \[D=0\] then the zeros are real and equal
Now, we need to find a quadratic polynomial with no real zeros
Let us assume the quadratic polynomial as
\[\Rightarrow a{{x}^{2}}+bx+c\]
Now, for this quadratic polynomial to have no real zeros from the above conditions its discriminant should be less than zero given by
\[\Rightarrow \sqrt{{{b}^{2}}-4ac}<0\]
Now, by squaring on both sides we get,
\[\Rightarrow {{b}^{2}}-4ac<0\]
Now, this can be further written as
\[\Rightarrow {{b}^{2}}<4ac\]
Now, let us consider the a, b, c values which satisfy the above condition
\[a=1,b=2,c=3\]
These values satisfy the condition.
Now, on substituting these values in the quadratic polynomial considered we get,
\[\therefore {{x}^{2}}+2x+3\]
Hence, the zeros of \[{{x}^{2}}+2x+3\] are not real.

Note:
Instead of assuming the variables and then consider the values according to the condition we can also solve this by taking the zeros in the direct formula to be some no real numbers and then solve further to get the values of a, b, c.
It is important to note that after squaring the inequality does not change and remains the same. It is also to be noted that we can take any values that satisfy the condition not something in particular.