Question

How do you solve $ {x^{\dfrac{3}{2}}} - 2{x^{\dfrac{3}{4}}} + 1 = 0 $ ?

Answer 1

Hint: To solve this problem. We should know about polynomial equations and algebraic identity.
Polynomial equation: The equation formed with variable and exponent and coefficient are called as polynomial equations.
As we have an algebraic identity: $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ .

Complete step by step solution:
To solve this first we try to simplify the given equation $ {x^{\dfrac{2}{3}}} - 2{x^{\dfrac{3}{4}}} + 1 = 0 $ .
Let’s take $ t = {x^{\dfrac{3}{4}}} $ .
We can write it as,
  $ {t^2} = {x^{\left( {\dfrac{3}{4}} \right)2}} = {x^{\left( {\dfrac{3}{2}} \right)}} $
Let's keep back in the given equation. We get,
  $ {t^2} - 2t + 1 = 0 $
By using algebraic identity $ {(a - b)^2} = {a^2} - 2ab + {b^2} $
So, we can write it as;
  $ {\left( {t - 1} \right)^2} = 0 $
So, the roots of the given quadratic equation we will, $ {\left( {t - 1} \right)^2} = 0 $
  $ \Rightarrow \left( {t - 1} \right) = 0 $
  $ \Rightarrow t = 1 $
As we have $ t = 1 $ .
So, we can write it as $ t = {x^{\dfrac{3}{4}}} = 1 $
If $ x \geqslant 0 $ then;
 \[x = {x^1} = {x^{\dfrac{3}{4}.\dfrac{4}{3}}} = {\left( {{x^{\dfrac{3}{4}}}} \right)^{\dfrac{4}{3}}} = {1^{\dfrac{4}{3}}} = 1\]
So, $ x = 1 $
There will only be real roots.

Note: Quadratic equation is used in daily life. It is used in calculating area, calculating a product’s profit or determining the speed of an object. It is used to rise and fall of profit from selling goods, the decrease and increase in the amount of time it take to run a mile based on your age, and so on