Question

How do you solve \[6x + 29 = \dfrac{5}{x}\]?

Answer 1

Hint: Here we have a linear equation with one variable. In the given problem we need to solve this for ‘x’. We can solve this using the transposition method. After simplifying this we will obtain a quadratic equation, hence we will have two solutions for ‘x’. We can solve the quadratic equation by the factorization method or formula method.

Complete step-by-step solution:
Given, \[6x + 29 = \dfrac{5}{x}\].
To simplify this easily we first multiply ‘x’ on both sides we have
\[x\left( {6x + 29} \right) = 5\]
\[6{x^2} + 29x = 5\]
\[ \Rightarrow 6{x^2} + 29x - 5 = 0\]
Thus we have a quadratic equation.
Now consider the expression \[6{x^2} + 29x - 5\]. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We have \[a = 6\], \[b = 29\] and \[c = - 5\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that\[{b_1} \times {b_2} = ac\] and\[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = 30\] and \[{b_2} = - 1\]. Because \[{b_1} \times {b_2} = - 30\] \[(a \times c)\] and \[{b_1} + {b_2} = 29(b)\].
Now we write \[6{x^2} + 29x - 5\] as,
\[ \Rightarrow 6{x^2} + 29x - 5 = 6{x^2} + 30x - x - 5 = 0\]
\[ \Rightarrow 6{x^2} + 30x - x - 5 = 0\]
Taking ‘6x’ common in the first two terms and taking -1 common in the remaining two terms we have,
\[ \Rightarrow 6x(x + 5) - 1(x + 5) = 0\]
Again taking \[(x + 5)\] common we have,
\[ \Rightarrow (x + 5)(6x - 1) = 0\].
Now using zero product principle we have
\[ \Rightarrow x + 5 = 0\] and \[6x - 1 = 0\]
\[ \Rightarrow x = - 5\] and \[6x = 1\]
\[ \Rightarrow x = - 5\] and \[x = \dfrac{1}{6}\]. This is the required result.

Note: In case if we are unable to solve the obtained quadratic equation using factorization, we can use the quadratic formula to solve it. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. We directly simplify the values of ‘a’, ‘b’, and ‘c’. We simplify it further to obtained the desired result. The above quadratic equation can be solved using the quadratic formula and we will have same answer in both the cases.