Question

Solve the equation $\left(5-2\sqrt6\right)^x+\left(5+2\sqrt6\right)^x$ =98 for $x$.

Answer 1

Hint: Initially, it may seem a difficult question but is really simple if observed properly. We will replace the term $5-2\sqrt{6\;}\;\mathrm{or}\;5+2\sqrt6\;$ by a variable. First, we have to find a relationship between these two terms. We can then solve the quadratic equation by a suitable method.
On multiplying $5-2\sqrt{6\;}\;\mathrm{and}\;5+2\sqrt6\;$, we find that-
$\left(5-2\sqrt6\right)\left(5+2\sqrt6\right)=\left(5^2-\left(2\sqrt6\right)^2\right)\\=25-6\times4=25-24=1\\\mathrm{Hence},\;\left(5-2\sqrt6\right)=\dfrac1{\left(5+2\sqrt6\right)}$

Complete step-by-step answer:

Now, let us assume $\left(5-2\sqrt6\right)^x$ as k, so $\left(5+2\sqrt6\right)^{\mathrm x}=\dfrac1{\mathrm k}.$ The equation can now be written as-
$\mathrm k+\dfrac1{\mathrm k}=98\\\mathrm k^2+1=98\mathrm k\\\mathrm k^2-98\mathrm k+1=0$
We will apply the quadratic formula to the equation which is-
$\dfrac{-\mathrm b\pm\sqrt{\mathrm b^2-4\mathrm{ac}}}{2\mathrm a}

\mathrm k=\dfrac{98\pm\sqrt{98^2-4}}2\\=\dfrac{98\pm\sqrt{9600}}2=\dfrac{98\pm40\sqrt6}2\\\mathrm k=49\pm20\sqrt6$
Now, we know the value of k in terms of x, so we can equate these values to find the value of x-
$\mathrm k=\left(5-2\sqrt6\right)^{\mathrm x}=\left(49\pm20\sqrt6\right)=\left(5\pm2\sqrt6\right)^2\\\mathrm{In}\;\mathrm{the}\;\mathrm{first}\;\mathrm{case},\\\left(5-2\sqrt6\right)^{\mathrm x}=\left(5-2\sqrt6\right)^2\\\mathrm x=2\\\mathrm{In}\;\mathrm{second}\;\mathrm{case},\\\left(5-2\sqrt6\right)^{\mathrm x}=\left(5+2\sqrt6\right)^2=\left(\dfrac1{\left(5-2\sqrt6\right)}\right)^2=\left(5-2\sqrt6\right)^{-2}\\\mathrm x=-2$
Hence, the values of x are -2 and 2.

Note: While finding the value of x using the value of k, it is important to take both the cases, to get two values of x, Students often forget one of the solutions and get only a partially correct answer. Also, it is important to find a relation between the two terms so that we can proceed with the equations.