Question

Find the roots of ${5^{\left( {x + 1} \right)}} + {5^{\left( {2 - x} \right)}} = {5^3} + 1$ by factorisation method.

Answer 1

Hint:Factorisation method is the technique of taking out the common terms from the equation and writing the remaining equation in brackets. Roots of an equation are the values which satisfy the equation and an equation can have more than one root. We will first take the factor common in all the terms outside the bracket and then simplify the equation. We will use some laws of exponents to form the quadratic equation and then solve the equation using factorization.

Complete step by step answer:
${5^{x + 1}} + {5^{2 - x}} = {5^3} + 1$
Calculating the cube and simplifying the right side of the equation, we get,
$ \Rightarrow {5^{x + 1}} + {5^{2 - x}} = 125 + 1 = 126$
We know the laws of exponents ${x^{n + m}} = {x^n} \times {x^m}$ and ${x^{n - m}} = \dfrac{{{x^n}}}{{{x^m}}}$.
Now, using the laws of exponents, we get,
$ \Rightarrow 5\left( {{5^x}} \right) + \dfrac{{{5^2}}}{{{5^x}}} = 126$
Taking the LCM of the rational terms, we get,
$ \Rightarrow \dfrac{{5\left( {{5^{2x}}} \right) + {5^2}}}{{{5^x}}} = 126$
Now, cross multiplying the terms of the equation, we get,
\[ \Rightarrow 5\left( {{5^{2x}}} \right) + {5^2} = 126\left( {{5^x}} \right)\]
Shifting all the terms to the left side of the equation, we get,
\[ \Rightarrow 5\left( {{5^{2x}}} \right) - 126\left( {{5^x}} \right) + 25 = 0\]

Now, let us assume ${5^x}$ as $t$. So, we get,
\[ \Rightarrow 5\left( {{t^2}} \right) - 126\left( t \right) + 25 = 0\]
Now, we solve the quadratic equation using factorization and splitting the middle terms method.
\[ \Rightarrow 5{t^2} - 125t - t + 25 = 0\]
\[ \Rightarrow 5t\left( {t - 25} \right) - \left( {t - 25} \right) = 0\]
Now, taking the common factor outside from all the terms, we get,
\[ \Rightarrow \left( {5t - 1} \right)\left( {t - 25} \right) = 0\]
Now, either \[\left( {5t - 1} \right) = 0\] or \[\left( {t - 25} \right) = 0\]
Either \[t = \dfrac{1}{5}\] or $t = 25$.
So, the values of t which satisfy the equation \[5{t^2} - 126t + 25 = 0\] are \[\dfrac{1}{5}\] and $25$.
Therefore, ${5^x} = 25$ or ${5^x} = \dfrac{1}{5}$
So, we get, ${5^x} = {5^2}$ or ${5^x} = {5^{ - 1}}$
Now, on comparing both sides of the equation, we get, $x = 2$ and $x = - 1$.

Note: We must know the standard form of a quadratic equation \[a{x^2} + bx + c = 0\]. There are various methods of solving any quadratic equation such as splitting the middle term, using quadratic formula and completing the square method. So, we can employ any of these methods to solve the equation formed in the question. We must take care of the calculations in order to be sure of the final answer.