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The minimum value of the expression $4{x^2} + 2x + 1 = 0$ is:
1) 1
2) $\dfrac{1}{2}$
3) $\dfrac{3}{4}$
4) $\dfrac{1}{3}$

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: The above given equation is a quadratic equation whose minimum value can be find out by checking the coefficient of x square and the standard equation for knowing whether the equation has minimum value or not;
If the coefficient of x2 is positive then it means the equation has positive value a> 0.
Using the above criteria we will solve the given problem.

Complete step by step answer:
Let’s discuss the condition to calculate the minimum value in more detail.
We will compare the given equation with the standard quadratic equation, which is given as,
$a{x^2} + bx + c = 0$
Equation given to us is;
$4{x^2} + 2x + 1 = 0$
After comparison we observe that,
a>0, which means that the function does not have maximum value but will have only minimum value.
Now, the calculation of minimum value is as follows;
$ \Rightarrow \dfrac{{ - {b^2} - 4ac}}{{4a}}$, using this formula we will calculate the minimum value.
$ \Rightarrow \dfrac{{ - ({2^2} - 4 \times 4 \times 1)}}{{4 \times 4}}$ (We have substituted the value of a, b and c)
$ \Rightarrow \dfrac{{12}}{{16}}$
$ \Rightarrow \dfrac{3}{4}$ (we obtain the value of the expression)

So, the correct answer is Option 3.

Note: We have another method to calculate the minimum and maximum value of a given function, which can be calculated by the method of differentiation. After performing the first derivative, we will calculate the value of the unknown and then substitute that value in the second derivative of the function, if the value after substituting in the second derivative the value is negative then the function has maximum value and if the function has positive value then the derivative has minimum value.