Question

Find the value of x from the following
\[3x + \dfrac{5}{{16x}} - 2 = 0\]

Answer 1

Hint: To solve a given equation, we first rearrange the equation to form a quadratic equation in one variable. Here the variable is x. After rearranging the equation, we find the value of x by factorization method or by quadratic formula. If factorization is not possible, then we use a quadratic formula. Here we are using both methods to find the values of x.

Complete step-by-step answer:
Given: An equation
\[3x + \dfrac{5}{{16x}} - 2 = 0\]
We can write above equation as
\[\dfrac{{48{x^2} + 5 - 32x}}{{16x}} = 0\]
\[ \Rightarrow 48{x^2} + 5 - 32x = 0\]
In the above quadratic equation we can use the factorization method. Or we use quadratic formulas.
Method 1: Factorization method
\[48{x^2} - 32x + 5 = 0\]
\[ \Rightarrow 48{x^2} - 12x - 20x + 5 = 0\]
\[ \Rightarrow 12x(4x - 1) - 5(4x - 1) = 0\]
\[ \Rightarrow (4x - 1)(12x - 5) = 0\]
We get
\[(4x - 1) = 0\]
\[\therefore x = \dfrac{1}{4}\]
and \[(12x - 5) = 0\]
\[\therefore x = \dfrac{5}{{12}}\]
Method II:
By quadratic formula: If \[a{x^2} + bx + c = 0\]is a quadratic equation, then quadratic formula is given by
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
For\[48{x^2} - 32x + 5 = 0\], \[a = 48.b = - 32,c = 5\]
we get, \[a = 48.b = - 32,c = 5\]
Substituting values of a.b and c in quadratic formula, we get
\[x = \dfrac{{ - ( - 32) \pm \sqrt {{{( - 32)}^2} - 4 \times 48 \times 5} }}{{2 \times 48}}\]
\[ \Rightarrow x = \dfrac{{32 \pm 8}}{{96}}\]
then we get two values of x as,
\[x = \dfrac{{32 + 8}}{{96}}and\dfrac{{32 - 8}}{{96}}\]
\[ \Rightarrow x = \dfrac{{40}}{{96}},\dfrac{{24}}{{96}}\]
\[\therefore x = \dfrac{5}{{12}},\dfrac{1}{{12}}\]

Hence, we get the value of x is \[\dfrac{5}{{12}}\& \dfrac{1}{{12}}\] by both methods.

Note: A quadratic equation in algebra is described by an equation that can be rearranged in standard form as \[A{x^2} + Bx + C = 0\]. Where x is unknown, which value is to be calculated in the equation. It is also called a variable. A,B and C represent known numbers or constants, Here \[a \ne 0\]. If\[a = 0\], then the equation is not quadratic, Then it is called linear.
The values of x that satisfy the equation are called solutions. Here the power of the equation is 2, so there are two values of x are obtained. These values are called roots or zeros of the expression.