Question

If $\sqrt {1 + \dfrac{{55}}{{729}}} = 1 + \dfrac{x}{{27}}$ , then find $x$ .

Answer 1

Hint: As the left hand side of the equation is in roots square on both sides and simplify the resulting equation. We will get a quadratic equation in terms of x. As we already know how to solve a quadratic equation to get x, we can use that particular formula and find the value of x.
The formula that will be used in this question will be:
The roots of the equation $a{x^2} + bx + c = 0$ are $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Complete step by step answer:
Given that the equation is $\sqrt {1 + \dfrac{{55}}{{729}}} = 1 + \dfrac{x}{{27}}$
By squaring on both sides of the given equation we get
                   ${\left( {\sqrt {1 + \dfrac{{55}}{{729}}} } \right)^2}$ = ${\left( {1 + \dfrac{x}{{27}}} \right)^2}$
We know that
${\left( {a + b} \right)^2}$ = ${a^2} + {b^2} + 2ab$ .
That is the square of the sum of two terms is expanded as the sum of squares of both terms and two times the product of them.
By applying the above formulae in the above equation we get
                           ${\left( {\sqrt {1 + \dfrac{{55}}{{729}}} } \right)^2}$ = ( 1+$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ )
 By simplifying the above equation, we get
      1 + $\dfrac{{55}}{{729}}$ = ( 1+$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ )
     $\dfrac{{729 + 55}}{{729}}$ = ( 1+$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ )
         $\dfrac{{784}}{{729}}$ = 1+$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$
         We can write this as
      1+$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ = $\dfrac{{784}}{{729}}$
By taking 1 to the right side of the equation we get
$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ = $\dfrac{{784}}{{729}}$ $ - $ 1
 By solving the equation we get
$\dfrac{{{x^2}}}{{{{\left( {27} \right)}^2}}}$ + $\dfrac{{2x}}{{27}}$ = $\dfrac{{784 - 729}}{{729}}$
By taking LCM on L.H.S we get the equation as
$\dfrac{{{x^2} + 2\left( {27} \right)x}}{{729}}$ = $\dfrac{{55}}{{729}}$
$\dfrac{{{x^2} + 54x}}{{729}}$ = $\dfrac{{55}}{{729}}$
 ${x^2} + 54x$ = 55
 ${x^2} + 54x$-55 = 0
We know that the roots of the equation $a{x^2} + bx + c = 0$ are $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
Which implies the roots of the above equation are \[\begin{gathered}
  x = \dfrac{{ - (54) \pm \sqrt {{{54}^2} - 4(1)( - 55)} }}{{2(1)}} \\
  \,\,\,\, = \dfrac{{ - (54) \pm \sqrt {{{54}^2} + 220} }}{{2(1)}} \\
  \,\,\,\, = \dfrac{{ - (54) \pm \sqrt {{{56}^2}} }}{2} \\
  \,\,\,\, = \dfrac{{ - 54 \pm 56}}{2} = 1\,\,or\, - 55 \\
\end{gathered} \]
So as the root of a value can be either positive or negative the solution for this question will be x = 1 or -55.

Note:
Remember that the root of any number can be positive or negative. Remember the formula for roots of quadratic equations and find the roots without any calculation mistakes. And check both sides while squaring both sides.