Question

Find for what value of \[a\] , equation \[{x^2} + 2x - \left( {{a^3} - 3a - 3} \right) = 0\] has real roots.

Answer 1

Hint: A quadratic equation will have real solution when \[D \geqslant 0\] , where \[D = {b^2} - 4ac\]
In this question we are given a quadratic equation as the degree of the given equation is two, so first, we will check the discriminant \[D\] of the equation whether the equation has real roots or not and then by substituting the values in the equation and then by solving them we will find the value of \[a\] , which is when substituted in the equation, the equation gives the real roots.

Complete step by step answer:
Given the equation \[{x^2} + 2x - \left( {{a^3} - 3a - 3} \right) = 0\] , since the degree of the equation is 2 so we can say the given equation is a quadratic equation, where
 \[
  a = 1 \\
  b = 2 \\
  c = - \left( {{a^3} - 3a - 3} \right) \\
 \]
Now we know for the equation to have real roots \[D \geqslant 0\] , where \[D = {b^2} - 4ac\]
So by substituting the values in the discriminant \[D\] , we get
 \[
  {b^2} - 4ac \geqslant 0 \\
   \Rightarrow {\left( 2 \right)^2} - 4\left( 1 \right)\left( { - \left( {{a^3} - 3a - 3} \right)} \right) \geqslant 0 \\
   \Rightarrow 4 + 4\left( {{a^3} - 3a - 3} \right) \geqslant 0 \\
   \Rightarrow 1 + 1\left( {{a^3} - 3a - 3} \right) \geqslant 0 \\
 \]
This can also be written as
 \[{a^3} - 3a - 2 \geqslant 0\]
Now we will find the factors of the equation, which can be written as
 \[\left( {a + 1} \right)\left( {{a^2} - a - 2} \right) \geqslant 0\left\{ {\because \left( {a + 1} \right) \times \left( {{a^2} - a - 2} \right) = {a^3} - 3a - 2} \right\}\]
By further solving this, we can write
 \[
  \left( {a + 1} \right)\left( {{a^2} - 2a - a - 2} \right) \geqslant 0 \\
   \Rightarrow \left( {a + 1} \right)\left( {a - 2} \right)\left( {a + 1} \right) \geqslant 0 \\
 \]
So,
 \[a = - 1,2\]
Hence we can say the value of \[a\] in the equation \[{x^2} + 2x - \left( {{a^3} - 3a - 3} \right) = 0\] is \[ - 1\] and \[2\]
Now if we substitute the value of \[a\] in the equation, the equation gives the real roots.

Note:
For a quadratic equation \[a{x^2} + bx + c = 0\] , its roots are generally found by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , since the degree of a quadratic equation is two, so a quadratic equation gives two roots.
Now if in quadratic equation discriminant \[D\] which is \[D = {b^2} - 4ac\]
 \[D > 0\] , Equation has two real solutions
 \[D = 0\] , Equation has one real solution
 \[D < 0\] , Equation has complex solutions