Courses
Courses for Kids
Free study material
Free LIVE classes
More

If one of the roots of ${x^2} + f\left( a \right)x + a = 0$ is equal to the third power of the other for all real a, then$a.{\text{ }}$ The domain of the real value function $f$ is the set of non-negative real numbers.$b.{\text{ }}f\left( x \right) = - {x^{\dfrac{1}{4}}}\left( {1 + {x^{\dfrac{1}{2}}}} \right) \\ c.{\text{ }}f\left( x \right) = - {x^{\dfrac{1}{4}}} + {x^{\dfrac{3}{2}}} \\ d.{\text{ None of these}} \\$

Last updated date: 17th Mar 2023
Total views: 307.8k
Views today: 5.87k
Verified
307.8k+ views
Hint: - Use sum of the roots $= \dfrac{{ - {\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$, and product of roots $= \dfrac{{{\text{constant term}}}}{{{\text{Coefficient of }}{x^2}}}$

Given quadratic equation ${x^2} + f\left( a \right)x + a = 0$
Let $\alpha$and $\beta$be the roots of the equation.
As you know sum of the roots $= \dfrac{{ - {\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$
And product of roots $= \dfrac{{{\text{constant term}}}}{{{\text{Coefficient of }}{x^2}}}$
$\Rightarrow \alpha + \beta = \dfrac{{ - f\left( a \right)}}{1} = - f\left( a \right).........\left( 1 \right), \\ \alpha \beta = \dfrac{a}{1} = a...............\left( 2 \right) \\$
Now it is given that one root is the cube of other
$\Rightarrow {\alpha ^3} = \beta$
From equation 2, considering real values of $\alpha$ we get
${\alpha ^3}\alpha = a \\ \Rightarrow {\alpha ^4} = a \\ \Rightarrow \alpha = \pm {a^{\dfrac{1}{4}}} \\$
For positive value of $\alpha$
$\Rightarrow \alpha = + {a^{\dfrac{1}{4}}}.............\left( 3 \right)$
Therefore from equation 1
$\alpha + \beta = - f\left( a \right) \\ \Rightarrow \alpha + {\alpha ^3} = - f\left( a \right) \\$
Now, from equation 3
$\Rightarrow {a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}} = - f\left( a \right) \\ \Rightarrow f\left( a \right) = - \left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}}} \right) \\ \Rightarrow f\left( a \right) = - {a^{\dfrac{1}{4}}}\left( {1 + {a^{\dfrac{1}{2}}}} \right) \\$
Now, in place of $a$ substitute $x$ in the above equation.
$\Rightarrow f\left( x \right) = - {x^{\dfrac{1}{4}}}\left( {1 + {x^{\dfrac{1}{2}}}} \right)$
Hence, option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that for a quadratic equation always remember the sum of roots and product of roots which is stated above, then simplify according to the given condition we will get the required answer.