Question

Solve the following equation:
\[{a^2} + ax < 1 - x\]
(a) None of these
(b) Cannot be determined
(c) Does not exist.
(d) \[1 - a\]

Answer 1

Hint: The given problem revolves around the concept's algebraic solution to solve the equation to get the desired value (or, roots). As a result, by simply separating the variable part and constant part on one side that is LH.S. and R.H.S. respectively, the desired root is obtained.

Complete answer:Since, we have given that
The given equation is,
\[{a^2} + ax < 1 - x\]
(which is the conditional equation showing the L.H.S. that is the left hand side part of the equation is less than that of the right hand side that is R.H.S. shown by the mathematical sign ‘\[ < \]’ of the respective given equation)
Since, the highest power of the equation is \[2\], it implies that the given equation is or is in the ‘quadratic’ form respectively.
Hence, by solving the equation algebraically that is taking the R.H.S. part to the L.H.S., we get
\[{a^2} + ax - 1 + x < 0\]
As a result, taking the constant on one side and variable terms on another side, we get
\[ax + x < 1 - {a^2}\]
Solving the equation mathematically, we get
\[x\left( {a + 1} \right) < 1 - {a^2}\]
\[x < \dfrac{{{1^2} - {a^2}}}{{a + 1}}\] … (\[\because {1^2} = 1\])
\[x < \dfrac{{{1^2} - {a^2}}}{{1 + a}}\] … (\[\because \] By associative law i.e. if \[a = b\] then \[b = a\])
Since, to get the value to its simplest
Multiplying and dividing by its conjugate that is \[1 - a\], we get
\[x < \dfrac{{{1^2} - {a^2}}}{{1 + a}} \times \dfrac{{1 - a}}{{1 - a}}\]
\[x < \dfrac{{\left( {{1^2} - {a^2}} \right) \times \left( {1 - a} \right)}}{{\left( {1 + a} \right) \times \left( {1 - a} \right)}}\]
We know that, \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\], we get
\[x < \dfrac{{\left( {{1^2} - {a^2}} \right) \times \left( {1 - a} \right)}}{{\left( {{1^2} - {a^2}} \right)}}\]
Solving the equation, we get
\[ \Rightarrow x < \left( {1 - a} \right)\] Where, ‘\[a\]’ is any constant!
Hence, the root of the given equation is \[1 - a\] respectively.
\[\therefore \]Option (d) is correct.

Note:
 One must be able to know basic properties while solving such equations such as separating the variables and constants, etc. Also, remember all the rules of indices like \[{a^m} \times {a^n} = {a^{m + n}}\], \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], etc. properties or, laws which includes (in terms of) in mathematics such as Associative law \[a + \left( {b + c} \right) = \left( {a + b} \right) + c\], Distributive law \[a \times \left( {b + c} \right) = a \times b + a \times c\], commutative law \[a + b = b + a\], so as to be sure of our final answer.