Question

Let one root of the equation \[{{x}^{2}}+\ell x+m=0\] is square of the other root. If \[m\in R\] then
A. \[\ell \in \left( -\infty ,\dfrac{1}{4} \right] \cup \left\{ 1 \right\}\]
B. \[\ell \in \left( -\infty ,0 \right] \]
C. \[\ell \in \left( -\infty ,\dfrac{1}{9} \right] \]
D. \[\ell \in \left( \dfrac{1}{4},1 \right] \]

Answer 1

Hint: We have given the quadratic equation i.e. \[{{x}^{2}}+\ell x+m=0\] and considering the given condition in the question we need to let the roots of the given quadratic equation be \[\alpha \] and \[{{\alpha }^{2}}\] . Then using the concepts of roots of the quadratic equation, we need to find the sum and the product of the roots of the quadratic equation. Later solving for the value of the root, substituting the value of \[\alpha ={{m}^{\dfrac{1}{3}}}\] . We will get the quadratic equation and then we know that if the roots are real then the value of discriminant is greater than and equal to 0 i.e. \[D={{b}^{2}}-4ac\ge 0\] . Substituting the value in the discriminant and solving will give us the required answer.
Formula used:
Standard form of quadratic equation i.e. \[a{{x}^{2}}+bx+c\] , where \[\alpha \] and \[\beta \] are the roots of the equation
Thus,
 Sum of the roots = \[\alpha +\beta =\dfrac{-b}{a}\]
Product of the roots = \[\alpha \times \beta =\dfrac{c}{a}\]

Complete step-by-step answer:
We have given that,
We have given the quadratic equation i.e.
 \[{{x}^{2}}+\ell x+m=0\]
Now,
It is given that,
One root of the equation \[{{x}^{2}}+\ell x+m=0\] is square of the other root.
Let the roots of the given quadratic equation be \[\alpha \] and \[{{\alpha }^{2}}\] .
Therefore,
As we know that,
Sum of the roots = \[\alpha +{{a}^{2}}=-\dfrac{\ell }{1}=-\ell \]
And
Product of the roots = \[\alpha \times {{\alpha }^{2}}={{\alpha }^{3}}=\dfrac{m}{1}=m\]
Now,
We have,
 \[\Rightarrow {{\alpha }^{3}}=m\ \Rightarrow \alpha ={{m}^{\dfrac{1}{3}}}\] ------- (1)
As we have,
 \[\alpha +{{a}^{2}}=-\ell \]
Substituting the value of \[\alpha ={{m}^{\dfrac{1}{3}}}\] , we will get
 \[{{m}^{\dfrac{1}{3}}}+{{m}^{\dfrac{2}{3}}}=-\ell \]
Adding \[\ell \] to both the sides of the equation,
 \[{{m}^{\dfrac{1}{3}}}+{{m}^{\dfrac{2}{3}}}+\ell =0\]
It can be rewrite as,
 \[{{\left( {{m}^{\dfrac{1}{3}}} \right)}^{2}}+{{m}^{\dfrac{1}{3}}}+\ell =0\]
As we can observe that,
It is the form of a quadratic equation, whose roots must be real.
So,
If the roots are real then the value of discriminant is greater than and equal to 0 i.e.
 \[D={{b}^{2}}-4ac\ge 0\]
Here,
 \[b=1,\ a=1\ and\ c=\ell \]
Substituting the values,
 \[\Rightarrow {{b}^{2}}-4ac\ge 0\]
 \[\Rightarrow {{1}^{2}}-4\left( 1 \right)\ell \ge 0\ \]
 \[\Rightarrow 1-4\ell \ge 0\ \]
 \[\Rightarrow 4\ell \le 1\]
Therefore,
 \[\Rightarrow \ell \le \dfrac{1}{4}\]
Thus,
 \[\ell \in \left( -\infty ,\dfrac{1}{4} \right] \cup \left\{ 1 \right\}\]
So, the correct answer is “Option A”.

Note: For solving these types of questions students should need to know the concepts of the roots of the quadratic equation.
Standard form of quadratic equation i.e. \[a{{x}^{2}}+bx+c\] , where \[\alpha \] and \[\beta \] are the roots of the equation;
Thus, Sum of the roots = \[\alpha +\beta =\dfrac{-b}{a}\] and Product of the roots = \[\alpha \times \beta =\dfrac{c}{a}\]
An equation is said to be quadratic if the highest value of degree is equal to 2. Roots of the quadratic equation are the values of x, when substituted in the equation satisfies the equation. A root of the polynomial is a value where the polynomial is equal to zero. To solve these types of questions, we need to write the given polynomial in the standard form i.e. decreasing order.