Question

Solve for \[x\]:
\[\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0\]

Answer 1

Hint: In the given problem, we have to solve for x, which means we have to find the value of \[x\].
Now, in this question, there is a polynomial which is quadratic and the coefficients of all the terms are irrational.
Now, we can find the value of \[x\] by using different methods like completing the squares, discriminant method, factorization method. In this prob, we are going to use the factorization method.

Complete step-by-step answer:
The given problem is:
\[\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0\]
On comparing it with \[a{x^2} + bx + c = 0\] which is the general equation in case of quadratic polynomial, we get:
\[a = \sqrt 3 ,b = - 2\sqrt 2 ,c = - 2\sqrt 3 \]
In case of factorization method, the coefficient ‘\[b\]’ should be written as sum of the terms which, after getting multiplied, form the product of ‘\[a\]’ and ‘\[c\]’.
$\Rightarrow$ product of \[a\] and \[c = \sqrt 3 \times \left( { - 2\sqrt 3 } \right)\]
\[ = - 2 \times \sqrt 3 \times \sqrt 3 \]
\[ = - 2 \times 3\]
\[ = - 6\]
So, on factoring \[\left( { - 6} \right)\] in the form of \[\sqrt 2 \] as \[b = - 2\sqrt 2 \],
We get, \[\left( { - 3\sqrt 2 } \right) \times \left( {\sqrt 2 } \right)\]
Or, \[\left( { - \sqrt 2 } \right) \times \left( {3\sqrt 2 } \right)\]
So, we will choose \[\left( { - 3\sqrt 2 } \right) \times \left( {\sqrt 2 } \right)\] as after addition, they will become \[\left( { - 2\sqrt 2 } \right)\] $\Rightarrow$ b.
Hence,
\[\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0\]
$\Rightarrow$ \[\sqrt 3 {x^2} - 3\sqrt 2 x + \sqrt 2 x - 2\sqrt 3 = 0\]
$\Rightarrow$ \[\sqrt 3 {x^2} - \sqrt 3 \sqrt 3 \sqrt 2 x + \sqrt 2 x - \sqrt 2 \sqrt 2 \sqrt 3 = 0\]
Because \[\sqrt a \sqrt a = {\left( {\sqrt a } \right)^2} = a\]
Hence, \[\sqrt 2 \sqrt 2 = {\left( {\sqrt 2 } \right)^2} = 2\]
Similarly, \[\sqrt 3 \sqrt 3 = {\left( {\sqrt 3 } \right)^2} = 3\]
Now, by taking \[\sqrt 3 x\] common from first two terms and \[\sqrt 2 \] common from last two terms, we get:
\[\sqrt 3 x\left( {x - \sqrt 3 \sqrt 2 } \right) + \sqrt 2 \left( {x - \sqrt 2 \sqrt 3 } \right) = 0\]
Also, \[\sqrt a \sqrt b = \sqrt {ab} \].
Therefore, \[\sqrt 2 \sqrt 3 = \sqrt {2 \times 3} = \sqrt 6 \]
$\Rightarrow$ \[\sqrt 3 x\left( {x - \sqrt 6 } \right) + \sqrt 2 \left( {x - 6} \right) = 0\]
$\Rightarrow$ \[\left( {\sqrt 3 x + \sqrt 2 } \right) + \left( {x - \sqrt 6 } \right) = 0\]
Now, either \[\sqrt 3 x + \sqrt 2 = 0\]
$\Rightarrow$ \[\sqrt 3 x = - \sqrt 2 \]
$\Rightarrow$ \[x = \dfrac{{ - \sqrt 2 }}{{\sqrt 3 }}\]
or, \[x - \sqrt 6 = 0\]
$\Rightarrow$ \[x = \sqrt 6 \]
Hence, \[x\] can have two values, \[\sqrt 6 \] or \[\dfrac{{ - \sqrt 2 }}{{\sqrt 3 }}\].

Note: The irrational numbers are those real numbers that are not rational i.e irrational numbers can’t be expressed as the ratio of two integers. For example, \[\sqrt 7 ,\sqrt 2 \] etc.
Now, the given problem was having the irrational numbers as the coefficients. So, we used the factorization method to solve the problem. Discriminant method could also be used where discriminant \[D = {b^2} - 4ac\] and \[x\] can be calculated be the formula:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Hence, by using any of the methods, we can get the solution i.e \[x = \sqrt 6 \] or \[x = \dfrac{{ - \sqrt 2 }}{3}\] as the polynomial is quadratic, at most two solutions of \[x\] are possible.