Question

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

Answer 1

Hint: The equation is the form of algebraic equation or expression. The algebraic equation or expression is a combination of variable and constant. By using the simple arithmetic operations, we can solve the above equation. Hence, we can find the solution for the variable x.

Complete step-by-step solution:
The equation is an algebraic equation or expression. It is a combination of variables and constants. In algebraic equations we have arithmetic operations also. Here in this question, we have to find the value of x.
Now consider the \[x + 5 = \dfrac{6}{x}\]
The RHS of the above equation is in the form of fraction.
Now multiply the equation by x we have
\[ \Rightarrow x \times (x + 5) = \dfrac{6}{x} \times x\]
On multiplying and simplifying we get
\[ \Rightarrow {x^2} + 5x = 6\]
Take 6 to LHS
\[ \Rightarrow {x^2} + 5x - 6 = 0\]
When we compare the given equation to the general form of equation that is \[a{x^2} + bx + c\] we have a = 1, b = 5 and c = -6.
Substituting these values in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] we get
\[ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4(1)( - 6)} }}{{2(1)}}\]
On simplifying we get
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 + 24} }}{2}\]
On further simplification we get
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {49} }}{2}\]
The number 49 is a perfect square. So the square root 49 is 7 and it is written as
\[ \Rightarrow x = \dfrac{{ - 5 \pm 7}}{2}\]
Therefore we have
\[ \Rightarrow x = \dfrac{{ - 5 + 7}}{2} = \dfrac{2}{2} = 1\] and \[x = \dfrac{{ - 5 - 7}}{2} = \dfrac{{ - 12}}{2} = - 6\]
Hence we have obtained the value of x
Therefore x = 1 and x = -6
We can also verify the obtained answer is correct is not.
So now consider the equation
\[x + 5 = \dfrac{6}{x}\]
Substitute the value of x as 1 in the above equation
\[ \Rightarrow 1 + 5 = \dfrac{6}{1}\]
On simplifying we get
\[ \Rightarrow 6 = 6\]
Hence we obtained the LHS is equal to RHS.
Therefore x = 1 is the correct value.
Substitute the value of x as -6 in the above equation
\[ \Rightarrow - 6 + 5 = \dfrac{6}{{ - 6}}\]
On simplifying we get
\[ \Rightarrow - 1 = - 1\]
Hence we obtained the LHS is equal to RHS.
Therefore x = -6 is the correct value.
Therefore x = 1 and x = -6

Note: While solving the equation we shift or transform the terms either from LHS to RHS or from RHS to LHS we should take care of the sign. Because while shifting or transforming the terms the sign of the term will change. If we miss out the sign we may go wrong while finding the variable or solving.