Answer

Verified

306.9k+ views

**Hint:**We are given an equation in irrationals and we need to simplify it and find the value of the unknown variable $x$. We need to apply some trick involving the irrationals here. We will use the concept of inverses in irrationals and we will consider one variable as $y$ and then the other term will become $\dfrac{1}{y}$. After doing this we will get an equation in a different variable $y$ and then we will solve for that, then re-substitute it in the older variable to get the value.

**Complete step by step answer:**

Let $y=\left(5+2\sqrt{6}\right)^{x^2-3}$

Then, if we multiply the quantity inside bracket by 1 i.e. $\dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}$, then we get:

$\left(5+2\sqrt{6}\right)\times \dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}=\dfrac{25-24}{5-2\sqrt{6}}=\dfrac{1}{5-2\sqrt{6}}$

$\implies \dfrac{1}{y}=5-2\sqrt{6}$

So, the equation becomes:

$y+\dfrac{1}{y}=10$

$\implies y^2+1=10y$

$\implies y^2-10y+1=0$

Solving this equation using the discriminant method:

For any general quadratic equation:

$ay^2+by+c=0$, the roots of this equation will be given by:

$y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\dfrac{10\pm \sqrt{100-4}}{2\times 1}=\dfrac{10\pm \sqrt{96}}{2}$

$\implies y=\dfrac{10\pm 4\sqrt{6}}{2}$

$\implies y=5\pm 2\sqrt{6}$

Putting back the value of $y$ in the equation above:

$y=\left(5+2\sqrt{6}\right)^{x^2-3}$

$5+2\sqrt{6}=\left(5+2\sqrt{6}\right)^{x^2-3}$

Or we have:

$5-2\sqrt{6}=\left(5+2\sqrt{6}\right)^{x^2-3}$

$\implies \left(5+2\sqrt{6}\right)^1=\left(5+2\sqrt{6}\right)^{x^2-3}$, or

${{\left( 5-2\sqrt{6} \right)}^{-1}}={{\left( 5+2\sqrt{6} \right)}^{{{x}^{2}}-3}}$

$\implies x^2-3=1$ or $x^2-3=-1$

$\implies x^2=4$ or $x^2=2$

$\implies x=\pm 2$ or $x=\pm\sqrt{2}$

Hence, the value of $x$ has been found out.

**Note:**Do not forget the $\pm$ sign while writing for the solution of any quadratic equation because then you might miss some of the solutions. You should remember the fact that the quadratic equation has two solutions if the discriminant is non-zero, which is the case here. Moreover, while dealing with the irrationals make sure you make no calculation mistakes.

Recently Updated Pages

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

The branch of science which deals with nature and natural class 10 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE

Classify weak and strong ligands out of a OH b F c class 12 chemistry CBSE

Define Vant Hoff factor How is it related to the degree class 12 chemistry CBSE