Question

Find zeros of \[y = 4{x^2} + 1\]?

Answer 1

Hint: Since for any quadratic polynomial, the zeros are given when the polynomial is equated to $0$, we equate the quadratic polynomial to $0$. Use a method of determinant to solve for the value of $x$ from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation. Solve the value under the square root and write two zeros by separating plus and minus signs from the obtained answer.
• For a polynomial \[p(x) = a{x^2} + bx + c\], the zeros are given when we put \[p(x) = 0\]
• For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step answer:
We are given the quadratic equation \[4{x^2} + 1\] -----(1)
We know that the general quadratic equation is \[a{x^2} + bx + c = 0\] where ‘a’, ‘b’, and ‘c’ are constant values.
On comparing the quadratic equation in equation (1) with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 4,b = 0,c = 1\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - (0) \pm \sqrt {{{(0)}^2} - 4 \times 1 \times 4} }}{{2 \times 4}}\]
Square the terms under the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ \pm \sqrt { - 16} }}{{2 \times 4}}\]
We can write \[\sqrt { - 16} = 4\sqrt { - 1} \]. Substitute this value in the numerator of the fraction.
\[ \Rightarrow x = \dfrac{{ \pm 4\sqrt { - 1} }}{{2 \times 4}}\]
Cancel same factor i.e. 4 from numerator and denominator
\[ \Rightarrow x = \dfrac{{ \pm \sqrt { - 1} }}{2}\]
But we know that \[\sqrt { - 1} = i\]
So, roots become \[x = \dfrac{i}{2}\]and \[x = \dfrac{{ - i}}{2}\]
Since roots are imaginary, then we don’t have any zeros of the quadratic polynomial \[y = 4{x^2} + 1\]
\[\therefore \] There are no real zeros of the equation \[y = 4{x^2} + 1\].

Note: Alternate method:
We can use Discriminant method \[(D = {b^2} - 4ac)\] which stats that if:
\[D > 0\] then there are 2 distinct real zeros
\[D = 0\] then there are repeated real zeros
\[D < 0\] then there are no real zeros
On comparing the quadratic equation in equation (1) with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 4,b = 0,c = 1\]
Substitute the values in formula:
\[D = 0 - 4 \times 4 \times 1\]
i.e. \[D = - 16\]
Since \[D < 0\], there are no real zeros for the equation.