Question

How do you solve $4{x^2} + 4x + 1 > 0$?

Answer 1

Hint: In order to verify whether the given polynomial is positive or negative, we need to find its roots. If $a{x^2} + bx + c = 0$ is a quadratic polynomial, we find the roots using the formula $x = {{ - b \pm \sqrt D }}{{2a}}$, where D is discriminant which is found using the formula $D = \sqrt {{b^2} - 4ac} $. From the value of D we will get to know about the nature of the roots. We try to make the given polynomial positive by imposing the condition on the obtained roots.

Complete step by step solution:
Here we need to prove that $4{x^2} + 4x + 1 > 0$.
i.e. we need to prove that the given polynomial is positive.
Since here the variable is x, we try to impose conditions on x to obtain the polynomial as positive.
So let us consider the quadratic polynomial $4{x^2} + 4x + 1 = 0$.
If $a{x^2} + bx + c = 0$ is a quadratic polynomial, then we find the roots using the formula, $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ ……(1)
where D is discriminant of the quadratic polynomial which is found using the formula $D = \sqrt {{b^2} - 4ac} $ ……(2)
In the given problem, $a = 4,\,\,b = 4,\,\,c = 1$
First we find the value of discriminant D.
Using the equation (2), we get,
$D = \sqrt {{4^2} - 4 \times 4 \times 1} $
$ \Rightarrow D = \sqrt {16 - 16} $
$ \Rightarrow D = \sqrt 0 $
 $ \Rightarrow D = 0 $
Hence the roots are real and equal.
i.e. the polynomial has only one real root.
Now the root of the polynomial $4{x^2} + 4x + 1 = 0$ using the equation (1).
In equation (1), substituting the values of D, a, b we get,
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - 4 \pm \sqrt 0 }}{{2 \times 4}} $
$ \Rightarrow x = \dfrac{{ - 4 \pm 0}}{8}$
$ \Rightarrow x = \dfrac{{ - 4}}{8}$
$ \Rightarrow x = \dfrac{{ - 1}}{2}$
But $x = \dfrac{{ - 1}}{2}$ is not possible.
Which means that the given quadratic polynomial is always positive except at its roots $(\because a > 0)$.

Hence the quadratic polynomial $4{x^2} + 4x + 1 > 0$ if and only if $x \ne - \dfrac{1}{2}$.

Note: Here we can’t use the method of factorization to find the roots or values of x for this question. Most of the students write the coefficient of x as $2x + 2x$, but this is wrong as it will give us sum as $4x$ but it will not give us the product of terms to be coefficient of ${x^2}$. We need to keep in mind that the coefficient of x should be broken in such a way that its sum gives us the value of the coefficient x and its product gives the value of the coefficient of ${x^2}$.
It's also important to remember the formula to find the roots of the quadratic equation $a{x^2} + bx + c = 0$ which is given by $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, where D is discriminant given by $D = \sqrt {{b^2} - 4ac} $.
The value of discriminant can be positive, negative or zero. And it tells us the nature of the roots.